TITLE: Proving a convergence relationship between two sequences
QUESTION [6 upvotes]: Let $a_{n}$ a sequence of real numbers. Let $\sigma_n= \frac{a_1+a_2+...+a_n}{n}$. Suppose that $\lim_{n\to \infty} \sigma_n=A.$
Prove that
$$\lim_{n \to \infty}\frac{1}{\log n}  \sum_{j=1}^{n}\frac{a_j}{j}=A$$

REPLY [8 votes]: I'd start from Abel Summation Identity, 
$$\begin{align}\sum\limits_{j=1}^{n} \frac{a_j}{j} &= \sum\limits_{j=1}^{n-1}\left(\frac{1}{j} - \frac{1}{j+1}\right)(a_1+\cdots+a_j) + \frac{1}{n}(a_1+\cdots+a_n)\\&= \sum\limits_{j=1}^{n-1}\frac{\sigma_j}{j+1} + \sigma_n\end{align}$$
Since, $\displaystyle \lim_{n \to \infty} \sigma_n = A$ (exists) and hence $|\sigma_n|$ is bounded, (by, $C > 0$ say) we may argue that. 
$\displaystyle \left|\sum\limits_{j=1}^{n-1}\frac{\sigma_j}{j+1}\right| \le C\sum\limits_{j=1}^{n-1}\frac{1}{j+1} < C \log n$
Hence, the limit $\displaystyle \lim\limits_{n\to \infty} \frac{1}{\log n}\sum\limits_{j=1}^{n-1}\frac{\sigma_j}{j+1}$ (exists) and equals $B$ (say).
By applying the Stolz-Cesaro Theorem,
$$B = \lim\limits_{n\to \infty} \frac{\sigma_n/(n+1)}{\log (n+1) - \log n} = \lim\limits_{n\to \infty} \frac{\sigma_n}{\log \left(1+\frac{1}{n}\right)^{n+1}} = \lim\limits_{n\to \infty} \sigma_n = A$$