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Q.

Show that if $\lim _{n \rightarrow \infty} a_{n}=a$, then $$ \lim _{n \rightarrow \infty} \frac{1}{\ln n} \sum_{k=1}^{n} \frac{a_{k}}{k}=a . $$

I didn't find it in old posts, anyway I think Stolz theorem can be used.

$\lim _{n \rightarrow \infty} \frac{\sum_{k=1}^{n} \frac{a_{k}}{k}}{\ln n}=\lim _{n \rightarrow \infty} \frac{a_{n+1}}{\ln \left(1+\frac{1}{n}\right)^{n+1}}=a .$

Is my solution okay? You can give your approach. Thank you.

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Another way to use Landau notations $$a_n=a+o(1)$$ and the well-known fact that $$\sum_{k=1}^n \frac 1 k = \ln n +o(\ln n)$$ Then

$$\frac{1}{\ln n} \sum_{k=1}^{n} \frac{a_{k}}{k}=\frac{1}{\ln n} \sum_{k=1}^{n} \frac{a+o(1)}{k}=\frac{a}{\ln n} \sum_{k=1}^{n} \frac{1}{k} +o\left(\frac{1}{\ln n} \sum_{k=1}^{n} \frac{1}{k}\right)=a+o(1)$$

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