For a given sequence $a_1,a_2,\cdots,a_n$ if $\lim_{n \to \infty} a_n=a$,then the question is to find out the value of $$\lim_{n \to \infty} \frac{1}{\ln (n)} \sum_{k=1}^{n} \frac{a_k}{k}$$

I know that for finding out the limit of sequence we have to equate $a_n=a_{n+1}$ but I am not aware how i can use it here.Please help me in this regard.Thanks.


A sum in a limit where you only know about the underlying sequence rather than the sum itself is a good situation to try the Stolz-Cesaro Theorem. Assuming the limit exists, we have

$$\lim_{n \to \infty} \frac{\sum_{k=1}^n \frac{a_k}{k}}{\log(n)} = \lim_{n \to \infty} \frac{ \sum_{k=1}^{n+1} \frac{a_k}{k} - \sum_{k=1}^n \frac{a_k}{k}}{\log(n+1) - \log(n)} = \lim_{n \to \infty} \frac{a_{n+1}}{(n+1)\log \left( 1 + \frac{1}{n} \right)}$$

Now you just have to work out the denominator and distribute your limits.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.