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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.

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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.

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