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

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

Try: Using Stolz Method ,

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

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

Answer given is $a$,

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

  • 1
    $\begingroup$ A more general result is true. If $a_n\to L$ 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. $\endgroup$ – Paramanand Singh Nov 30 '18 at 2:58

By Stolz-Cesaro

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


$$\lim_{n\rightarrow \infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =\lim_{n\rightarrow \infty}\frac{\frac{a_{n+1}}{n+1}}{\ln (n+1)-\ln n}=\lim_{n\rightarrow \infty}\frac{\frac{n}{n+1}a_{n+1}}{\ln \left(1+\frac1n\right)^n} \to a$$


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.

  • $\begingroup$ I have a doubt why $\lim_{n\rightarrow \infty}a_{n+1}=a.$ please explain me $\endgroup$ – DXT Nov 29 '18 at 14:52
  • $\begingroup$ @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) $\epsilon$. $\endgroup$ – Sorin Tirc Nov 29 '18 at 15:30

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