value of Then $lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$












0















For a given sequence $a_{1},a_{2},a_{3},....,a_{n}$.



If $lim_{nrightarrow infty} a_{n} = a.$ Then $displaystyle lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$




Try: Using Stolz Method ,



Let $displaystyle frac{a_{n}}{b_{n}} = lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$



Then $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = lim_{nrightarrow infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})}=a_{n+1}$$



Answer given is $a$,



Could some help me Why $a_{n+1}=a_{n}$ for $nrightarrow infty,$ thanks






limits







share|cite|improve this question


















  • 1




    A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
    – Paramanand Singh
    Nov 30 at 2:58


















0















For a given sequence $a_{1},a_{2},a_{3},....,a_{n}$.



If $lim_{nrightarrow infty} a_{n} = a.$ Then $displaystyle lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$




Try: Using Stolz Method ,



Let $displaystyle frac{a_{n}}{b_{n}} = lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$



Then $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = lim_{nrightarrow infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})}=a_{n+1}$$



Answer given is $a$,



Could some help me Why $a_{n+1}=a_{n}$ for $nrightarrow infty,$ thanks






limits







share|cite|improve this question


















  • 1




    A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
    – Paramanand Singh
    Nov 30 at 2:58
















0












0








0


1






For a given sequence $a_{1},a_{2},a_{3},....,a_{n}$.



If $lim_{nrightarrow infty} a_{n} = a.$ Then $displaystyle lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$




Try: Using Stolz Method ,



Let $displaystyle frac{a_{n}}{b_{n}} = lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$



Then $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = lim_{nrightarrow infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})}=a_{n+1}$$



Answer given is $a$,



Could some help me Why $a_{n+1}=a_{n}$ for $nrightarrow infty,$ thanks






limits







share|cite|improve this question














For a given sequence $a_{1},a_{2},a_{3},....,a_{n}$.



If $lim_{nrightarrow infty} a_{n} = a.$ Then $displaystyle lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$




Try: Using Stolz Method ,



Let $displaystyle frac{a_{n}}{b_{n}} = lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$



Then $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = lim_{nrightarrow infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})}=a_{n+1}$$



Answer given is $a$,



Could some help me Why $a_{n+1}=a_{n}$ for $nrightarrow infty,$ thanks






limits




limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 at 14:22









D Tiwari

5,3492630




5,3492630








  • 1




    A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
    – Paramanand Singh
    Nov 30 at 2:58
















  • 1




    A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
    – Paramanand Singh
    Nov 30 at 2:58










1




1




A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
– Paramanand Singh
Nov 30 at 2:58






A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
– Paramanand Singh
Nov 30 at 2:58












2 Answers
2






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2














By Stolz-Cesaro



$$frac{a_{n}}{b_{n}} = lim_{nrightarrow infty} frac{sum^{n}_{k=1}frac{a_{k}}{k}}{ln (n)}$$



then



$$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =lim_{nrightarrow infty}frac{frac{a_{n+1}}{n+1}}{ln (n+1)-ln n}=lim_{nrightarrow infty}frac{frac{n}{n+1}a_{n+1}}{ln left(1+frac1nright)^n} to a$$






share|cite|improve this answer





























    1














    What the author of the solution is using is that $lim_{n to infty}a_{n+1} = lim_{n to infty}a_n$.



    Therefore, we may rewrite the last step in a more rigorous way as



    $lim_{n to infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})} = frac{lim_{n to infty}a_{n+1}}{lim_{n to infty}ln(1+frac{1}{n})^{n+1}} = frac{a}{log e} = a$, where I also used the fact that $ lim_{n to infty}(1+frac{1}{n})^{n+1} = e$



    It doesn't have to be the case that $a_{n+1} = a_n$ for the solution to work.






    share|cite|improve this answer





















    • I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
      – D Tiwari
      Nov 29 at 14:52












    • @DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
      – Sorin Tirc
      Nov 29 at 15:30













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    2














    By Stolz-Cesaro



    $$frac{a_{n}}{b_{n}} = lim_{nrightarrow infty} frac{sum^{n}_{k=1}frac{a_{k}}{k}}{ln (n)}$$



    then



    $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =lim_{nrightarrow infty}frac{frac{a_{n+1}}{n+1}}{ln (n+1)-ln n}=lim_{nrightarrow infty}frac{frac{n}{n+1}a_{n+1}}{ln left(1+frac1nright)^n} to a$$






    share|cite|improve this answer


























      2














      By Stolz-Cesaro



      $$frac{a_{n}}{b_{n}} = lim_{nrightarrow infty} frac{sum^{n}_{k=1}frac{a_{k}}{k}}{ln (n)}$$



      then



      $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =lim_{nrightarrow infty}frac{frac{a_{n+1}}{n+1}}{ln (n+1)-ln n}=lim_{nrightarrow infty}frac{frac{n}{n+1}a_{n+1}}{ln left(1+frac1nright)^n} to a$$






      share|cite|improve this answer
























        2












        2








        2






        By Stolz-Cesaro



        $$frac{a_{n}}{b_{n}} = lim_{nrightarrow infty} frac{sum^{n}_{k=1}frac{a_{k}}{k}}{ln (n)}$$



        then



        $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =lim_{nrightarrow infty}frac{frac{a_{n+1}}{n+1}}{ln (n+1)-ln n}=lim_{nrightarrow infty}frac{frac{n}{n+1}a_{n+1}}{ln left(1+frac1nright)^n} to a$$






        share|cite|improve this answer












        By Stolz-Cesaro



        $$frac{a_{n}}{b_{n}} = lim_{nrightarrow infty} frac{sum^{n}_{k=1}frac{a_{k}}{k}}{ln (n)}$$



        then



        $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =lim_{nrightarrow infty}frac{frac{a_{n+1}}{n+1}}{ln (n+1)-ln n}=lim_{nrightarrow infty}frac{frac{n}{n+1}a_{n+1}}{ln left(1+frac1nright)^n} to a$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 at 14:58









        gimusi

        1




        1























            1














            What the author of the solution is using is that $lim_{n to infty}a_{n+1} = lim_{n to infty}a_n$.



            Therefore, we may rewrite the last step in a more rigorous way as



            $lim_{n to infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})} = frac{lim_{n to infty}a_{n+1}}{lim_{n to infty}ln(1+frac{1}{n})^{n+1}} = frac{a}{log e} = a$, where I also used the fact that $ lim_{n to infty}(1+frac{1}{n})^{n+1} = e$



            It doesn't have to be the case that $a_{n+1} = a_n$ for the solution to work.






            share|cite|improve this answer





















            • I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
              – D Tiwari
              Nov 29 at 14:52












            • @DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
              – Sorin Tirc
              Nov 29 at 15:30


















            1














            What the author of the solution is using is that $lim_{n to infty}a_{n+1} = lim_{n to infty}a_n$.



            Therefore, we may rewrite the last step in a more rigorous way as



            $lim_{n to infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})} = frac{lim_{n to infty}a_{n+1}}{lim_{n to infty}ln(1+frac{1}{n})^{n+1}} = frac{a}{log e} = a$, where I also used the fact that $ lim_{n to infty}(1+frac{1}{n})^{n+1} = e$



            It doesn't have to be the case that $a_{n+1} = a_n$ for the solution to work.






            share|cite|improve this answer





















            • I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
              – D Tiwari
              Nov 29 at 14:52












            • @DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
              – Sorin Tirc
              Nov 29 at 15:30
















            1












            1








            1






            What the author of the solution is using is that $lim_{n to infty}a_{n+1} = lim_{n to infty}a_n$.



            Therefore, we may rewrite the last step in a more rigorous way as



            $lim_{n to infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})} = frac{lim_{n to infty}a_{n+1}}{lim_{n to infty}ln(1+frac{1}{n})^{n+1}} = frac{a}{log e} = a$, where I also used the fact that $ lim_{n to infty}(1+frac{1}{n})^{n+1} = e$



            It doesn't have to be the case that $a_{n+1} = a_n$ for the solution to work.






            share|cite|improve this answer












            What the author of the solution is using is that $lim_{n to infty}a_{n+1} = lim_{n to infty}a_n$.



            Therefore, we may rewrite the last step in a more rigorous way as



            $lim_{n to infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})} = frac{lim_{n to infty}a_{n+1}}{lim_{n to infty}ln(1+frac{1}{n})^{n+1}} = frac{a}{log e} = a$, where I also used the fact that $ lim_{n to infty}(1+frac{1}{n})^{n+1} = e$



            It doesn't have to be the case that $a_{n+1} = a_n$ for the solution to work.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 29 at 14:36









            Sorin Tirc

            94710




            94710












            • I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
              – D Tiwari
              Nov 29 at 14:52












            • @DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
              – Sorin Tirc
              Nov 29 at 15:30




















            • I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
              – D Tiwari
              Nov 29 at 14:52












            • @DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
              – Sorin Tirc
              Nov 29 at 15:30


















            I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
            – D Tiwari
            Nov 29 at 14:52






            I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
            – D Tiwari
            Nov 29 at 14:52














            @DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
            – Sorin Tirc
            Nov 29 at 15:30






            @DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
            – Sorin Tirc
            Nov 29 at 15:30




















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