Infinite Series polynomials











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This is a homework problem.



Let P and Q be polynomials of degree $k$ and $m$.
Assume $Q(n) neq 0 $ for all $n in mathbb{N} $



Prove the series



$sum_{n=1}^{infty} {P(n)}/{Q(n)}$



Is convergent if $m geq k+2$ and divergent if $m leq k+1 $



From my understanding it would converge when $ m > k$ And therefore i can't figure out how it should be divergent when $m leq k+1$ because then $m=k+1$ and $m>k$










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    This is a homework problem.



    Let P and Q be polynomials of degree $k$ and $m$.
    Assume $Q(n) neq 0 $ for all $n in mathbb{N} $



    Prove the series



    $sum_{n=1}^{infty} {P(n)}/{Q(n)}$



    Is convergent if $m geq k+2$ and divergent if $m leq k+1 $



    From my understanding it would converge when $ m > k$ And therefore i can't figure out how it should be divergent when $m leq k+1$ because then $m=k+1$ and $m>k$










    share|cite|improve this question
























      up vote
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      This is a homework problem.



      Let P and Q be polynomials of degree $k$ and $m$.
      Assume $Q(n) neq 0 $ for all $n in mathbb{N} $



      Prove the series



      $sum_{n=1}^{infty} {P(n)}/{Q(n)}$



      Is convergent if $m geq k+2$ and divergent if $m leq k+1 $



      From my understanding it would converge when $ m > k$ And therefore i can't figure out how it should be divergent when $m leq k+1$ because then $m=k+1$ and $m>k$










      share|cite|improve this question













      This is a homework problem.



      Let P and Q be polynomials of degree $k$ and $m$.
      Assume $Q(n) neq 0 $ for all $n in mathbb{N} $



      Prove the series



      $sum_{n=1}^{infty} {P(n)}/{Q(n)}$



      Is convergent if $m geq k+2$ and divergent if $m leq k+1 $



      From my understanding it would converge when $ m > k$ And therefore i can't figure out how it should be divergent when $m leq k+1$ because then $m=k+1$ and $m>k$







      sequences-and-series polynomials






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      asked Nov 25 at 19:12









      Winther

      227




      227






















          3 Answers
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          accepted










          Write the polynomials as
          begin{align}
          P(x) &= a_kx^k+dots+a_1x+a_0 \[4px]
          Q(x) &= b_mx^n+dots+b_1x+b_0
          end{align}

          with $a_kne0$ and $b_mne0$.



          Then you can rewrite
          $$
          frac{P(n)}{Q(n)}=
          frac{n^k}{n^m}
          frac{a_k+dfrac{a_{k-1}}{n}+dots+dfrac{a_1}{n^{k-1}}+dfrac{a_0}{n^k}}
          {b_m+dfrac{b_{m-1}}{n}+dots+dfrac{b_1}{n^{m-1}}+dfrac{b_0}{n^m}}
          $$

          Note that the limit of the big fraction is $a_k/b_m=c$. If $c<0$, we can substitute $P(n)$ with $-P(n)$, without affecting the convergence or divergence of the series, because from some point on, all terms of the series have the same sign as $c$.



          Thus we can assume $c>0$. Then there exists $N$ such that, for every $n>N$, all terms $P(n)/Q(n)$ have the same sign and
          $$
          0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}
          $$

          Hence the given series converges if and only if
          $$
          sum_{n}frac{n^k}{n^m}
          $$

          converges.



          If $mle k$ this series doesn't converge, because the general term doesn't have zero limit. If $m=k+1$ this is the harmonic series, which does not converge.



          If $m>k+1$ (that is, $mge k+2$) then the series converges by comparison with $1/n^2$.






          share|cite|improve this answer





















          • Could you elaborate on how you go from $0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}$ to the series converges if and if only $sum_{n}frac{n^k}{n^m}$
            – Winther
            Nov 26 at 1:01








          • 1




            @ChristianWinther If $kne0$, then the series $sum_n a_n$ converges if and only if $sum_n (ka_n)$ converges. If $0le a_nle b_n$ and $sum_n b_n$ converges, then $sum_n a_n$ converges.
            – egreg
            Nov 26 at 8:54


















          up vote
          0
          down vote













          For $n$ big enough, $P(n)/Q(n)$ has always the same sign, so, WLOG, we can assume that we have a series with positive terms(*).



          Let $P(n)=p_0+p_1n+cdots+p_k n^k$. Because $P$ is of degree $k$, $p_kne0$.



          Let $Q(n)=q_0+q_1n+cdots+q_m n^m$. Because $Q$ is of degree $m$, $q_mne0$.



          Now, $$lim_{ntoinfty}frac{P(n)/Q(n)}{1/n^{m-k}}= lim_{ntoinfty}frac{n^{m-k} P(n)}{Q(n)}=lim_{ntoinfty}frac{p_0n^{m-k}+p_1n^{m-k+1}+cdots+p_kn^m}{q_0+q_1n+cdots q_mn^m}=frac{p_0}{q_0}.$$ So your series has the same character as $sum frac{1}{n^{m-k}}$ wich is convergente $iff$ $m-k>1$.



          But because $m$ and $k$ are both integers, $m-k>1$ $iff$ $m-kge 2$.



          EDIT (*):
          If $p_kcdot q_m>0$, then for $n$ big enough, $P(n)/Q(n)>0$ because both $P(n)$ and $Q(n)$ tends simultaneously to $+infty$ or $-infty$, when $ntoinfty$.



          If $p_kcdot q_m<0$, then $P(n)to+infty$ and $Q(n)to-infty$ ($ntoinfty$) or viceversa. Then $P(n)/Q(n)<0$ for $n$ big enough and then you only have to work with $-P(n)/Q(n)$.






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            The idea is that if $mgeq {k+2}$, then
            $$
            frac{P(n)}{Q(n)}sim frac{C}{n^{m-k}}
            $$

            where $f(n)sim g(n)$ means $f(n)/g(n)to 1$ as $ntoinfty$ for some constant $C$. The limit comparison test gives us the result. If $mleq k$ you should be able to show the terms of the series don't go to zero. If $m=k+1$, you should be able to apply limit comparison test by deducing that
            $$
            frac{P(n)}{Q(n)}sim frac{D}{n}.
            $$

            for some constant $D$.






            share|cite|improve this answer























            • That's almost correct -- you forgot the leading coefficients of $P$ and $Q$ in the ratio.
              – Clement C.
              Nov 25 at 19:22










            • But if $m=k+1$ with your explanation then it still converges to $0$, but the question asks me to prove it is divergent.
              – Winther
              Nov 25 at 19:47












            • Is it maybe a fault in the question or am I missing something?
              – Winther
              Nov 25 at 19:47










            • @ChristianWinther the sequence $frac1n$ converges to $0$ but the partial sums $sum_{n=1}^N frac{1}{n}$ diverges like $log N$.
              – achille hui
              Nov 25 at 22:13












            • @achillehui Thank you, i had forgotten about the harmonic series diverging
              – Winther
              Nov 25 at 22:17











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            3 Answers
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            3 Answers
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            active

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            up vote
            2
            down vote



            accepted










            Write the polynomials as
            begin{align}
            P(x) &= a_kx^k+dots+a_1x+a_0 \[4px]
            Q(x) &= b_mx^n+dots+b_1x+b_0
            end{align}

            with $a_kne0$ and $b_mne0$.



            Then you can rewrite
            $$
            frac{P(n)}{Q(n)}=
            frac{n^k}{n^m}
            frac{a_k+dfrac{a_{k-1}}{n}+dots+dfrac{a_1}{n^{k-1}}+dfrac{a_0}{n^k}}
            {b_m+dfrac{b_{m-1}}{n}+dots+dfrac{b_1}{n^{m-1}}+dfrac{b_0}{n^m}}
            $$

            Note that the limit of the big fraction is $a_k/b_m=c$. If $c<0$, we can substitute $P(n)$ with $-P(n)$, without affecting the convergence or divergence of the series, because from some point on, all terms of the series have the same sign as $c$.



            Thus we can assume $c>0$. Then there exists $N$ such that, for every $n>N$, all terms $P(n)/Q(n)$ have the same sign and
            $$
            0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}
            $$

            Hence the given series converges if and only if
            $$
            sum_{n}frac{n^k}{n^m}
            $$

            converges.



            If $mle k$ this series doesn't converge, because the general term doesn't have zero limit. If $m=k+1$ this is the harmonic series, which does not converge.



            If $m>k+1$ (that is, $mge k+2$) then the series converges by comparison with $1/n^2$.






            share|cite|improve this answer





















            • Could you elaborate on how you go from $0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}$ to the series converges if and if only $sum_{n}frac{n^k}{n^m}$
              – Winther
              Nov 26 at 1:01








            • 1




              @ChristianWinther If $kne0$, then the series $sum_n a_n$ converges if and only if $sum_n (ka_n)$ converges. If $0le a_nle b_n$ and $sum_n b_n$ converges, then $sum_n a_n$ converges.
              – egreg
              Nov 26 at 8:54















            up vote
            2
            down vote



            accepted










            Write the polynomials as
            begin{align}
            P(x) &= a_kx^k+dots+a_1x+a_0 \[4px]
            Q(x) &= b_mx^n+dots+b_1x+b_0
            end{align}

            with $a_kne0$ and $b_mne0$.



            Then you can rewrite
            $$
            frac{P(n)}{Q(n)}=
            frac{n^k}{n^m}
            frac{a_k+dfrac{a_{k-1}}{n}+dots+dfrac{a_1}{n^{k-1}}+dfrac{a_0}{n^k}}
            {b_m+dfrac{b_{m-1}}{n}+dots+dfrac{b_1}{n^{m-1}}+dfrac{b_0}{n^m}}
            $$

            Note that the limit of the big fraction is $a_k/b_m=c$. If $c<0$, we can substitute $P(n)$ with $-P(n)$, without affecting the convergence or divergence of the series, because from some point on, all terms of the series have the same sign as $c$.



            Thus we can assume $c>0$. Then there exists $N$ such that, for every $n>N$, all terms $P(n)/Q(n)$ have the same sign and
            $$
            0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}
            $$

            Hence the given series converges if and only if
            $$
            sum_{n}frac{n^k}{n^m}
            $$

            converges.



            If $mle k$ this series doesn't converge, because the general term doesn't have zero limit. If $m=k+1$ this is the harmonic series, which does not converge.



            If $m>k+1$ (that is, $mge k+2$) then the series converges by comparison with $1/n^2$.






            share|cite|improve this answer





















            • Could you elaborate on how you go from $0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}$ to the series converges if and if only $sum_{n}frac{n^k}{n^m}$
              – Winther
              Nov 26 at 1:01








            • 1




              @ChristianWinther If $kne0$, then the series $sum_n a_n$ converges if and only if $sum_n (ka_n)$ converges. If $0le a_nle b_n$ and $sum_n b_n$ converges, then $sum_n a_n$ converges.
              – egreg
              Nov 26 at 8:54













            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Write the polynomials as
            begin{align}
            P(x) &= a_kx^k+dots+a_1x+a_0 \[4px]
            Q(x) &= b_mx^n+dots+b_1x+b_0
            end{align}

            with $a_kne0$ and $b_mne0$.



            Then you can rewrite
            $$
            frac{P(n)}{Q(n)}=
            frac{n^k}{n^m}
            frac{a_k+dfrac{a_{k-1}}{n}+dots+dfrac{a_1}{n^{k-1}}+dfrac{a_0}{n^k}}
            {b_m+dfrac{b_{m-1}}{n}+dots+dfrac{b_1}{n^{m-1}}+dfrac{b_0}{n^m}}
            $$

            Note that the limit of the big fraction is $a_k/b_m=c$. If $c<0$, we can substitute $P(n)$ with $-P(n)$, without affecting the convergence or divergence of the series, because from some point on, all terms of the series have the same sign as $c$.



            Thus we can assume $c>0$. Then there exists $N$ such that, for every $n>N$, all terms $P(n)/Q(n)$ have the same sign and
            $$
            0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}
            $$

            Hence the given series converges if and only if
            $$
            sum_{n}frac{n^k}{n^m}
            $$

            converges.



            If $mle k$ this series doesn't converge, because the general term doesn't have zero limit. If $m=k+1$ this is the harmonic series, which does not converge.



            If $m>k+1$ (that is, $mge k+2$) then the series converges by comparison with $1/n^2$.






            share|cite|improve this answer












            Write the polynomials as
            begin{align}
            P(x) &= a_kx^k+dots+a_1x+a_0 \[4px]
            Q(x) &= b_mx^n+dots+b_1x+b_0
            end{align}

            with $a_kne0$ and $b_mne0$.



            Then you can rewrite
            $$
            frac{P(n)}{Q(n)}=
            frac{n^k}{n^m}
            frac{a_k+dfrac{a_{k-1}}{n}+dots+dfrac{a_1}{n^{k-1}}+dfrac{a_0}{n^k}}
            {b_m+dfrac{b_{m-1}}{n}+dots+dfrac{b_1}{n^{m-1}}+dfrac{b_0}{n^m}}
            $$

            Note that the limit of the big fraction is $a_k/b_m=c$. If $c<0$, we can substitute $P(n)$ with $-P(n)$, without affecting the convergence or divergence of the series, because from some point on, all terms of the series have the same sign as $c$.



            Thus we can assume $c>0$. Then there exists $N$ such that, for every $n>N$, all terms $P(n)/Q(n)$ have the same sign and
            $$
            0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}
            $$

            Hence the given series converges if and only if
            $$
            sum_{n}frac{n^k}{n^m}
            $$

            converges.



            If $mle k$ this series doesn't converge, because the general term doesn't have zero limit. If $m=k+1$ this is the harmonic series, which does not converge.



            If $m>k+1$ (that is, $mge k+2$) then the series converges by comparison with $1/n^2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 25 at 21:39









            egreg

            175k1383198




            175k1383198












            • Could you elaborate on how you go from $0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}$ to the series converges if and if only $sum_{n}frac{n^k}{n^m}$
              – Winther
              Nov 26 at 1:01








            • 1




              @ChristianWinther If $kne0$, then the series $sum_n a_n$ converges if and only if $sum_n (ka_n)$ converges. If $0le a_nle b_n$ and $sum_n b_n$ converges, then $sum_n a_n$ converges.
              – egreg
              Nov 26 at 8:54


















            • Could you elaborate on how you go from $0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}$ to the series converges if and if only $sum_{n}frac{n^k}{n^m}$
              – Winther
              Nov 26 at 1:01








            • 1




              @ChristianWinther If $kne0$, then the series $sum_n a_n$ converges if and only if $sum_n (ka_n)$ converges. If $0le a_nle b_n$ and $sum_n b_n$ converges, then $sum_n a_n$ converges.
              – egreg
              Nov 26 at 8:54
















            Could you elaborate on how you go from $0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}$ to the series converges if and if only $sum_{n}frac{n^k}{n^m}$
            – Winther
            Nov 26 at 1:01






            Could you elaborate on how you go from $0<frac{c}{2}frac{n^k}{n^m}<frac{P(n)}{Q(n)}<frac{3c}{2}frac{n^k}{n^m}$ to the series converges if and if only $sum_{n}frac{n^k}{n^m}$
            – Winther
            Nov 26 at 1:01






            1




            1




            @ChristianWinther If $kne0$, then the series $sum_n a_n$ converges if and only if $sum_n (ka_n)$ converges. If $0le a_nle b_n$ and $sum_n b_n$ converges, then $sum_n a_n$ converges.
            – egreg
            Nov 26 at 8:54




            @ChristianWinther If $kne0$, then the series $sum_n a_n$ converges if and only if $sum_n (ka_n)$ converges. If $0le a_nle b_n$ and $sum_n b_n$ converges, then $sum_n a_n$ converges.
            – egreg
            Nov 26 at 8:54










            up vote
            0
            down vote













            For $n$ big enough, $P(n)/Q(n)$ has always the same sign, so, WLOG, we can assume that we have a series with positive terms(*).



            Let $P(n)=p_0+p_1n+cdots+p_k n^k$. Because $P$ is of degree $k$, $p_kne0$.



            Let $Q(n)=q_0+q_1n+cdots+q_m n^m$. Because $Q$ is of degree $m$, $q_mne0$.



            Now, $$lim_{ntoinfty}frac{P(n)/Q(n)}{1/n^{m-k}}= lim_{ntoinfty}frac{n^{m-k} P(n)}{Q(n)}=lim_{ntoinfty}frac{p_0n^{m-k}+p_1n^{m-k+1}+cdots+p_kn^m}{q_0+q_1n+cdots q_mn^m}=frac{p_0}{q_0}.$$ So your series has the same character as $sum frac{1}{n^{m-k}}$ wich is convergente $iff$ $m-k>1$.



            But because $m$ and $k$ are both integers, $m-k>1$ $iff$ $m-kge 2$.



            EDIT (*):
            If $p_kcdot q_m>0$, then for $n$ big enough, $P(n)/Q(n)>0$ because both $P(n)$ and $Q(n)$ tends simultaneously to $+infty$ or $-infty$, when $ntoinfty$.



            If $p_kcdot q_m<0$, then $P(n)to+infty$ and $Q(n)to-infty$ ($ntoinfty$) or viceversa. Then $P(n)/Q(n)<0$ for $n$ big enough and then you only have to work with $-P(n)/Q(n)$.






            share|cite|improve this answer

























              up vote
              0
              down vote













              For $n$ big enough, $P(n)/Q(n)$ has always the same sign, so, WLOG, we can assume that we have a series with positive terms(*).



              Let $P(n)=p_0+p_1n+cdots+p_k n^k$. Because $P$ is of degree $k$, $p_kne0$.



              Let $Q(n)=q_0+q_1n+cdots+q_m n^m$. Because $Q$ is of degree $m$, $q_mne0$.



              Now, $$lim_{ntoinfty}frac{P(n)/Q(n)}{1/n^{m-k}}= lim_{ntoinfty}frac{n^{m-k} P(n)}{Q(n)}=lim_{ntoinfty}frac{p_0n^{m-k}+p_1n^{m-k+1}+cdots+p_kn^m}{q_0+q_1n+cdots q_mn^m}=frac{p_0}{q_0}.$$ So your series has the same character as $sum frac{1}{n^{m-k}}$ wich is convergente $iff$ $m-k>1$.



              But because $m$ and $k$ are both integers, $m-k>1$ $iff$ $m-kge 2$.



              EDIT (*):
              If $p_kcdot q_m>0$, then for $n$ big enough, $P(n)/Q(n)>0$ because both $P(n)$ and $Q(n)$ tends simultaneously to $+infty$ or $-infty$, when $ntoinfty$.



              If $p_kcdot q_m<0$, then $P(n)to+infty$ and $Q(n)to-infty$ ($ntoinfty$) or viceversa. Then $P(n)/Q(n)<0$ for $n$ big enough and then you only have to work with $-P(n)/Q(n)$.






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                For $n$ big enough, $P(n)/Q(n)$ has always the same sign, so, WLOG, we can assume that we have a series with positive terms(*).



                Let $P(n)=p_0+p_1n+cdots+p_k n^k$. Because $P$ is of degree $k$, $p_kne0$.



                Let $Q(n)=q_0+q_1n+cdots+q_m n^m$. Because $Q$ is of degree $m$, $q_mne0$.



                Now, $$lim_{ntoinfty}frac{P(n)/Q(n)}{1/n^{m-k}}= lim_{ntoinfty}frac{n^{m-k} P(n)}{Q(n)}=lim_{ntoinfty}frac{p_0n^{m-k}+p_1n^{m-k+1}+cdots+p_kn^m}{q_0+q_1n+cdots q_mn^m}=frac{p_0}{q_0}.$$ So your series has the same character as $sum frac{1}{n^{m-k}}$ wich is convergente $iff$ $m-k>1$.



                But because $m$ and $k$ are both integers, $m-k>1$ $iff$ $m-kge 2$.



                EDIT (*):
                If $p_kcdot q_m>0$, then for $n$ big enough, $P(n)/Q(n)>0$ because both $P(n)$ and $Q(n)$ tends simultaneously to $+infty$ or $-infty$, when $ntoinfty$.



                If $p_kcdot q_m<0$, then $P(n)to+infty$ and $Q(n)to-infty$ ($ntoinfty$) or viceversa. Then $P(n)/Q(n)<0$ for $n$ big enough and then you only have to work with $-P(n)/Q(n)$.






                share|cite|improve this answer












                For $n$ big enough, $P(n)/Q(n)$ has always the same sign, so, WLOG, we can assume that we have a series with positive terms(*).



                Let $P(n)=p_0+p_1n+cdots+p_k n^k$. Because $P$ is of degree $k$, $p_kne0$.



                Let $Q(n)=q_0+q_1n+cdots+q_m n^m$. Because $Q$ is of degree $m$, $q_mne0$.



                Now, $$lim_{ntoinfty}frac{P(n)/Q(n)}{1/n^{m-k}}= lim_{ntoinfty}frac{n^{m-k} P(n)}{Q(n)}=lim_{ntoinfty}frac{p_0n^{m-k}+p_1n^{m-k+1}+cdots+p_kn^m}{q_0+q_1n+cdots q_mn^m}=frac{p_0}{q_0}.$$ So your series has the same character as $sum frac{1}{n^{m-k}}$ wich is convergente $iff$ $m-k>1$.



                But because $m$ and $k$ are both integers, $m-k>1$ $iff$ $m-kge 2$.



                EDIT (*):
                If $p_kcdot q_m>0$, then for $n$ big enough, $P(n)/Q(n)>0$ because both $P(n)$ and $Q(n)$ tends simultaneously to $+infty$ or $-infty$, when $ntoinfty$.



                If $p_kcdot q_m<0$, then $P(n)to+infty$ and $Q(n)to-infty$ ($ntoinfty$) or viceversa. Then $P(n)/Q(n)<0$ for $n$ big enough and then you only have to work with $-P(n)/Q(n)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 at 19:20









                Tito Eliatron

                1,317622




                1,317622






















                    up vote
                    0
                    down vote













                    The idea is that if $mgeq {k+2}$, then
                    $$
                    frac{P(n)}{Q(n)}sim frac{C}{n^{m-k}}
                    $$

                    where $f(n)sim g(n)$ means $f(n)/g(n)to 1$ as $ntoinfty$ for some constant $C$. The limit comparison test gives us the result. If $mleq k$ you should be able to show the terms of the series don't go to zero. If $m=k+1$, you should be able to apply limit comparison test by deducing that
                    $$
                    frac{P(n)}{Q(n)}sim frac{D}{n}.
                    $$

                    for some constant $D$.






                    share|cite|improve this answer























                    • That's almost correct -- you forgot the leading coefficients of $P$ and $Q$ in the ratio.
                      – Clement C.
                      Nov 25 at 19:22










                    • But if $m=k+1$ with your explanation then it still converges to $0$, but the question asks me to prove it is divergent.
                      – Winther
                      Nov 25 at 19:47












                    • Is it maybe a fault in the question or am I missing something?
                      – Winther
                      Nov 25 at 19:47










                    • @ChristianWinther the sequence $frac1n$ converges to $0$ but the partial sums $sum_{n=1}^N frac{1}{n}$ diverges like $log N$.
                      – achille hui
                      Nov 25 at 22:13












                    • @achillehui Thank you, i had forgotten about the harmonic series diverging
                      – Winther
                      Nov 25 at 22:17















                    up vote
                    0
                    down vote













                    The idea is that if $mgeq {k+2}$, then
                    $$
                    frac{P(n)}{Q(n)}sim frac{C}{n^{m-k}}
                    $$

                    where $f(n)sim g(n)$ means $f(n)/g(n)to 1$ as $ntoinfty$ for some constant $C$. The limit comparison test gives us the result. If $mleq k$ you should be able to show the terms of the series don't go to zero. If $m=k+1$, you should be able to apply limit comparison test by deducing that
                    $$
                    frac{P(n)}{Q(n)}sim frac{D}{n}.
                    $$

                    for some constant $D$.






                    share|cite|improve this answer























                    • That's almost correct -- you forgot the leading coefficients of $P$ and $Q$ in the ratio.
                      – Clement C.
                      Nov 25 at 19:22










                    • But if $m=k+1$ with your explanation then it still converges to $0$, but the question asks me to prove it is divergent.
                      – Winther
                      Nov 25 at 19:47












                    • Is it maybe a fault in the question or am I missing something?
                      – Winther
                      Nov 25 at 19:47










                    • @ChristianWinther the sequence $frac1n$ converges to $0$ but the partial sums $sum_{n=1}^N frac{1}{n}$ diverges like $log N$.
                      – achille hui
                      Nov 25 at 22:13












                    • @achillehui Thank you, i had forgotten about the harmonic series diverging
                      – Winther
                      Nov 25 at 22:17













                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    The idea is that if $mgeq {k+2}$, then
                    $$
                    frac{P(n)}{Q(n)}sim frac{C}{n^{m-k}}
                    $$

                    where $f(n)sim g(n)$ means $f(n)/g(n)to 1$ as $ntoinfty$ for some constant $C$. The limit comparison test gives us the result. If $mleq k$ you should be able to show the terms of the series don't go to zero. If $m=k+1$, you should be able to apply limit comparison test by deducing that
                    $$
                    frac{P(n)}{Q(n)}sim frac{D}{n}.
                    $$

                    for some constant $D$.






                    share|cite|improve this answer














                    The idea is that if $mgeq {k+2}$, then
                    $$
                    frac{P(n)}{Q(n)}sim frac{C}{n^{m-k}}
                    $$

                    where $f(n)sim g(n)$ means $f(n)/g(n)to 1$ as $ntoinfty$ for some constant $C$. The limit comparison test gives us the result. If $mleq k$ you should be able to show the terms of the series don't go to zero. If $m=k+1$, you should be able to apply limit comparison test by deducing that
                    $$
                    frac{P(n)}{Q(n)}sim frac{D}{n}.
                    $$

                    for some constant $D$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 25 at 23:08

























                    answered Nov 25 at 19:16









                    Foobaz John

                    20.2k41250




                    20.2k41250












                    • That's almost correct -- you forgot the leading coefficients of $P$ and $Q$ in the ratio.
                      – Clement C.
                      Nov 25 at 19:22










                    • But if $m=k+1$ with your explanation then it still converges to $0$, but the question asks me to prove it is divergent.
                      – Winther
                      Nov 25 at 19:47












                    • Is it maybe a fault in the question or am I missing something?
                      – Winther
                      Nov 25 at 19:47










                    • @ChristianWinther the sequence $frac1n$ converges to $0$ but the partial sums $sum_{n=1}^N frac{1}{n}$ diverges like $log N$.
                      – achille hui
                      Nov 25 at 22:13












                    • @achillehui Thank you, i had forgotten about the harmonic series diverging
                      – Winther
                      Nov 25 at 22:17


















                    • That's almost correct -- you forgot the leading coefficients of $P$ and $Q$ in the ratio.
                      – Clement C.
                      Nov 25 at 19:22










                    • But if $m=k+1$ with your explanation then it still converges to $0$, but the question asks me to prove it is divergent.
                      – Winther
                      Nov 25 at 19:47












                    • Is it maybe a fault in the question or am I missing something?
                      – Winther
                      Nov 25 at 19:47










                    • @ChristianWinther the sequence $frac1n$ converges to $0$ but the partial sums $sum_{n=1}^N frac{1}{n}$ diverges like $log N$.
                      – achille hui
                      Nov 25 at 22:13












                    • @achillehui Thank you, i had forgotten about the harmonic series diverging
                      – Winther
                      Nov 25 at 22:17
















                    That's almost correct -- you forgot the leading coefficients of $P$ and $Q$ in the ratio.
                    – Clement C.
                    Nov 25 at 19:22




                    That's almost correct -- you forgot the leading coefficients of $P$ and $Q$ in the ratio.
                    – Clement C.
                    Nov 25 at 19:22












                    But if $m=k+1$ with your explanation then it still converges to $0$, but the question asks me to prove it is divergent.
                    – Winther
                    Nov 25 at 19:47






                    But if $m=k+1$ with your explanation then it still converges to $0$, but the question asks me to prove it is divergent.
                    – Winther
                    Nov 25 at 19:47














                    Is it maybe a fault in the question or am I missing something?
                    – Winther
                    Nov 25 at 19:47




                    Is it maybe a fault in the question or am I missing something?
                    – Winther
                    Nov 25 at 19:47












                    @ChristianWinther the sequence $frac1n$ converges to $0$ but the partial sums $sum_{n=1}^N frac{1}{n}$ diverges like $log N$.
                    – achille hui
                    Nov 25 at 22:13






                    @ChristianWinther the sequence $frac1n$ converges to $0$ but the partial sums $sum_{n=1}^N frac{1}{n}$ diverges like $log N$.
                    – achille hui
                    Nov 25 at 22:13














                    @achillehui Thank you, i had forgotten about the harmonic series diverging
                    – Winther
                    Nov 25 at 22:17




                    @achillehui Thank you, i had forgotten about the harmonic series diverging
                    – Winther
                    Nov 25 at 22:17


















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