Series convergence with parameter:












2















Determine whether the following series :
$$sum_{n=2}^infty frac {1+xn}{sqrt{n^2+n^6x}}, x in mathbb R^+_0$$
converges absolutely, conditionally or diverges.




I tried to estimate the series for $n > x$ using the following:
$$sum_{n=2}^infty frac {1+xn}{sqrt{n^2+n^6x}} leq sum_{n=2}^infty frac {2xn}{sqrt{n^6x}} = 2sqrt xsum_{n=2}^infty n^{-2}$$



Which would mean that the series is absolutely convergent for every $x in mathbb R^+_0$?



Is this the correct way to go about this or am I overlooking something?






calculus sequences-and-series proof-verification convergence







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    2















    Determine whether the following series :
    $$sum_{n=2}^infty frac {1+xn}{sqrt{n^2+n^6x}}, x in mathbb R^+_0$$
    converges absolutely, conditionally or diverges.




    I tried to estimate the series for $n > x$ using the following:
    $$sum_{n=2}^infty frac {1+xn}{sqrt{n^2+n^6x}} leq sum_{n=2}^infty frac {2xn}{sqrt{n^6x}} = 2sqrt xsum_{n=2}^infty n^{-2}$$



    Which would mean that the series is absolutely convergent for every $x in mathbb R^+_0$?



    Is this the correct way to go about this or am I overlooking something?






    calculus sequences-and-series proof-verification convergence







    share|cite|improve this question

























      2












      2








      2


      1






      Determine whether the following series :
      $$sum_{n=2}^infty frac {1+xn}{sqrt{n^2+n^6x}}, x in mathbb R^+_0$$
      converges absolutely, conditionally or diverges.




      I tried to estimate the series for $n > x$ using the following:
      $$sum_{n=2}^infty frac {1+xn}{sqrt{n^2+n^6x}} leq sum_{n=2}^infty frac {2xn}{sqrt{n^6x}} = 2sqrt xsum_{n=2}^infty n^{-2}$$



      Which would mean that the series is absolutely convergent for every $x in mathbb R^+_0$?



      Is this the correct way to go about this or am I overlooking something?






      calculus sequences-and-series proof-verification convergence







      share|cite|improve this question














      Determine whether the following series :
      $$sum_{n=2}^infty frac {1+xn}{sqrt{n^2+n^6x}}, x in mathbb R^+_0$$
      converges absolutely, conditionally or diverges.




      I tried to estimate the series for $n > x$ using the following:
      $$sum_{n=2}^infty frac {1+xn}{sqrt{n^2+n^6x}} leq sum_{n=2}^infty frac {2xn}{sqrt{n^6x}} = 2sqrt xsum_{n=2}^infty n^{-2}$$



      Which would mean that the series is absolutely convergent for every $x in mathbb R^+_0$?



      Is this the correct way to go about this or am I overlooking something?






      calculus sequences-and-series proof-verification convergence




      calculus sequences-and-series proof-verification convergence






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 3 '18 at 19:55









      J. Lastin

      1025




      1025






















          2 Answers
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          3














          Note that the series diverges if $x = 0$.



          You need a slight correction to your argument. For fixed $x > 0$, we have for all $n > 1/x$,



          $$ frac{1+xn}{sqrt{n^2 + n^6x}} leqslant frac{2nx}{n^3sqrt{x}} = frac{2sqrt{x}}{n^2},$$



          and by the comparison test we have pointwise convergence for all $x > 0$.



          The convergence is not uniform for $x in (0,infty)$. Note that



          $$sup_{x in [0,infty)}sum_{n=m+1}^{infty} frac{1+xn}{sqrt{n^2 + n^6x}} >sup_{x in [0,infty)}sum_{n=m+1}^{2m} frac{1}{nsqrt{1 + n^4x}} > sup_{x in [0,infty)}frac{m}{(2m)sqrt{1+(2m)^4x}} \ = sup_{x in [0,infty)}frac{1}{2sqrt{1+16m^4x}} geqslant underbrace{frac{1}{2sqrt{1+16m^4{m^{-4}}}}}_{text{ taking } x = m^{-4}} = frac{1}{2sqrt{17}} $$



          and the RHS does not converge to $0$ as $m to infty$.



          The convergence is uniform on any compact interval $[a,b]$ with $a > 0$. I will leave this for you to prove.






          share|cite|improve this answer































            0














            Perhaps you meant $n > frac{1}{x}$? Then
            $$
            sum_{n=2}^{infty} frac{1+xn}{sqrt{n^2 + n^6 x}} leq
            sum_{n=2}^{infty} frac{2nx}{sqrt{n^6 x}} =
            sum_{n=2}^{infty} frac{2 sqrt{x}}{n^2} =
            2sqrt{x} sum_{n=2}^{infty} frac{1}{n^2}
            $$
            which is convergent for each $x>0$.



            Edit: After I posted my answer I saw that @RRL had posted a more complete answer.






            share|cite|improve this answer























            • That was a typo. Yes, it is basically the same answer.
              – mlerma54
              Dec 3 '18 at 20:44










            • I didn't see your answer until I posted mine, otherwise I would have not posted it. Yours is also more complete since it discusses non uniform convergence.
              – mlerma54
              Dec 3 '18 at 20:47










            • No problem then,
              – RRL
              Dec 3 '18 at 20:47











            Your Answer





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            2 Answers
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            2 Answers
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            3














            Note that the series diverges if $x = 0$.



            You need a slight correction to your argument. For fixed $x > 0$, we have for all $n > 1/x$,



            $$ frac{1+xn}{sqrt{n^2 + n^6x}} leqslant frac{2nx}{n^3sqrt{x}} = frac{2sqrt{x}}{n^2},$$



            and by the comparison test we have pointwise convergence for all $x > 0$.



            The convergence is not uniform for $x in (0,infty)$. Note that



            $$sup_{x in [0,infty)}sum_{n=m+1}^{infty} frac{1+xn}{sqrt{n^2 + n^6x}} >sup_{x in [0,infty)}sum_{n=m+1}^{2m} frac{1}{nsqrt{1 + n^4x}} > sup_{x in [0,infty)}frac{m}{(2m)sqrt{1+(2m)^4x}} \ = sup_{x in [0,infty)}frac{1}{2sqrt{1+16m^4x}} geqslant underbrace{frac{1}{2sqrt{1+16m^4{m^{-4}}}}}_{text{ taking } x = m^{-4}} = frac{1}{2sqrt{17}} $$



            and the RHS does not converge to $0$ as $m to infty$.



            The convergence is uniform on any compact interval $[a,b]$ with $a > 0$. I will leave this for you to prove.






            share|cite|improve this answer




























              3














              Note that the series diverges if $x = 0$.



              You need a slight correction to your argument. For fixed $x > 0$, we have for all $n > 1/x$,



              $$ frac{1+xn}{sqrt{n^2 + n^6x}} leqslant frac{2nx}{n^3sqrt{x}} = frac{2sqrt{x}}{n^2},$$



              and by the comparison test we have pointwise convergence for all $x > 0$.



              The convergence is not uniform for $x in (0,infty)$. Note that



              $$sup_{x in [0,infty)}sum_{n=m+1}^{infty} frac{1+xn}{sqrt{n^2 + n^6x}} >sup_{x in [0,infty)}sum_{n=m+1}^{2m} frac{1}{nsqrt{1 + n^4x}} > sup_{x in [0,infty)}frac{m}{(2m)sqrt{1+(2m)^4x}} \ = sup_{x in [0,infty)}frac{1}{2sqrt{1+16m^4x}} geqslant underbrace{frac{1}{2sqrt{1+16m^4{m^{-4}}}}}_{text{ taking } x = m^{-4}} = frac{1}{2sqrt{17}} $$



              and the RHS does not converge to $0$ as $m to infty$.



              The convergence is uniform on any compact interval $[a,b]$ with $a > 0$. I will leave this for you to prove.






              share|cite|improve this answer


























                3












                3








                3






                Note that the series diverges if $x = 0$.



                You need a slight correction to your argument. For fixed $x > 0$, we have for all $n > 1/x$,



                $$ frac{1+xn}{sqrt{n^2 + n^6x}} leqslant frac{2nx}{n^3sqrt{x}} = frac{2sqrt{x}}{n^2},$$



                and by the comparison test we have pointwise convergence for all $x > 0$.



                The convergence is not uniform for $x in (0,infty)$. Note that



                $$sup_{x in [0,infty)}sum_{n=m+1}^{infty} frac{1+xn}{sqrt{n^2 + n^6x}} >sup_{x in [0,infty)}sum_{n=m+1}^{2m} frac{1}{nsqrt{1 + n^4x}} > sup_{x in [0,infty)}frac{m}{(2m)sqrt{1+(2m)^4x}} \ = sup_{x in [0,infty)}frac{1}{2sqrt{1+16m^4x}} geqslant underbrace{frac{1}{2sqrt{1+16m^4{m^{-4}}}}}_{text{ taking } x = m^{-4}} = frac{1}{2sqrt{17}} $$



                and the RHS does not converge to $0$ as $m to infty$.



                The convergence is uniform on any compact interval $[a,b]$ with $a > 0$. I will leave this for you to prove.






                share|cite|improve this answer














                Note that the series diverges if $x = 0$.



                You need a slight correction to your argument. For fixed $x > 0$, we have for all $n > 1/x$,



                $$ frac{1+xn}{sqrt{n^2 + n^6x}} leqslant frac{2nx}{n^3sqrt{x}} = frac{2sqrt{x}}{n^2},$$



                and by the comparison test we have pointwise convergence for all $x > 0$.



                The convergence is not uniform for $x in (0,infty)$. Note that



                $$sup_{x in [0,infty)}sum_{n=m+1}^{infty} frac{1+xn}{sqrt{n^2 + n^6x}} >sup_{x in [0,infty)}sum_{n=m+1}^{2m} frac{1}{nsqrt{1 + n^4x}} > sup_{x in [0,infty)}frac{m}{(2m)sqrt{1+(2m)^4x}} \ = sup_{x in [0,infty)}frac{1}{2sqrt{1+16m^4x}} geqslant underbrace{frac{1}{2sqrt{1+16m^4{m^{-4}}}}}_{text{ taking } x = m^{-4}} = frac{1}{2sqrt{17}} $$



                and the RHS does not converge to $0$ as $m to infty$.



                The convergence is uniform on any compact interval $[a,b]$ with $a > 0$. I will leave this for you to prove.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 3 '18 at 20:43

























                answered Dec 3 '18 at 20:34









                RRL

                49.2k42573




                49.2k42573























                    0














                    Perhaps you meant $n > frac{1}{x}$? Then
                    $$
                    sum_{n=2}^{infty} frac{1+xn}{sqrt{n^2 + n^6 x}} leq
                    sum_{n=2}^{infty} frac{2nx}{sqrt{n^6 x}} =
                    sum_{n=2}^{infty} frac{2 sqrt{x}}{n^2} =
                    2sqrt{x} sum_{n=2}^{infty} frac{1}{n^2}
                    $$
                    which is convergent for each $x>0$.



                    Edit: After I posted my answer I saw that @RRL had posted a more complete answer.






                    share|cite|improve this answer























                    • That was a typo. Yes, it is basically the same answer.
                      – mlerma54
                      Dec 3 '18 at 20:44










                    • I didn't see your answer until I posted mine, otherwise I would have not posted it. Yours is also more complete since it discusses non uniform convergence.
                      – mlerma54
                      Dec 3 '18 at 20:47










                    • No problem then,
                      – RRL
                      Dec 3 '18 at 20:47
















                    0














                    Perhaps you meant $n > frac{1}{x}$? Then
                    $$
                    sum_{n=2}^{infty} frac{1+xn}{sqrt{n^2 + n^6 x}} leq
                    sum_{n=2}^{infty} frac{2nx}{sqrt{n^6 x}} =
                    sum_{n=2}^{infty} frac{2 sqrt{x}}{n^2} =
                    2sqrt{x} sum_{n=2}^{infty} frac{1}{n^2}
                    $$
                    which is convergent for each $x>0$.



                    Edit: After I posted my answer I saw that @RRL had posted a more complete answer.






                    share|cite|improve this answer























                    • That was a typo. Yes, it is basically the same answer.
                      – mlerma54
                      Dec 3 '18 at 20:44










                    • I didn't see your answer until I posted mine, otherwise I would have not posted it. Yours is also more complete since it discusses non uniform convergence.
                      – mlerma54
                      Dec 3 '18 at 20:47










                    • No problem then,
                      – RRL
                      Dec 3 '18 at 20:47














                    0












                    0








                    0






                    Perhaps you meant $n > frac{1}{x}$? Then
                    $$
                    sum_{n=2}^{infty} frac{1+xn}{sqrt{n^2 + n^6 x}} leq
                    sum_{n=2}^{infty} frac{2nx}{sqrt{n^6 x}} =
                    sum_{n=2}^{infty} frac{2 sqrt{x}}{n^2} =
                    2sqrt{x} sum_{n=2}^{infty} frac{1}{n^2}
                    $$
                    which is convergent for each $x>0$.



                    Edit: After I posted my answer I saw that @RRL had posted a more complete answer.






                    share|cite|improve this answer














                    Perhaps you meant $n > frac{1}{x}$? Then
                    $$
                    sum_{n=2}^{infty} frac{1+xn}{sqrt{n^2 + n^6 x}} leq
                    sum_{n=2}^{infty} frac{2nx}{sqrt{n^6 x}} =
                    sum_{n=2}^{infty} frac{2 sqrt{x}}{n^2} =
                    2sqrt{x} sum_{n=2}^{infty} frac{1}{n^2}
                    $$
                    which is convergent for each $x>0$.



                    Edit: After I posted my answer I saw that @RRL had posted a more complete answer.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 3 '18 at 20:50

























                    answered Dec 3 '18 at 20:40









                    mlerma54

                    1,162138




                    1,162138












                    • That was a typo. Yes, it is basically the same answer.
                      – mlerma54
                      Dec 3 '18 at 20:44










                    • I didn't see your answer until I posted mine, otherwise I would have not posted it. Yours is also more complete since it discusses non uniform convergence.
                      – mlerma54
                      Dec 3 '18 at 20:47










                    • No problem then,
                      – RRL
                      Dec 3 '18 at 20:47


















                    • That was a typo. Yes, it is basically the same answer.
                      – mlerma54
                      Dec 3 '18 at 20:44










                    • I didn't see your answer until I posted mine, otherwise I would have not posted it. Yours is also more complete since it discusses non uniform convergence.
                      – mlerma54
                      Dec 3 '18 at 20:47










                    • No problem then,
                      – RRL
                      Dec 3 '18 at 20:47
















                    That was a typo. Yes, it is basically the same answer.
                    – mlerma54
                    Dec 3 '18 at 20:44




                    That was a typo. Yes, it is basically the same answer.
                    – mlerma54
                    Dec 3 '18 at 20:44












                    I didn't see your answer until I posted mine, otherwise I would have not posted it. Yours is also more complete since it discusses non uniform convergence.
                    – mlerma54
                    Dec 3 '18 at 20:47




                    I didn't see your answer until I posted mine, otherwise I would have not posted it. Yours is also more complete since it discusses non uniform convergence.
                    – mlerma54
                    Dec 3 '18 at 20:47












                    No problem then,
                    – RRL
                    Dec 3 '18 at 20:47




                    No problem then,
                    – RRL
                    Dec 3 '18 at 20:47


















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