Limit of $\frac{1}{\ln n}\sum_{j=1}^n\frac{x_j}{j}$ in two different cases. I have two related problems:
(1) For a sequence $\{x_n\}$ with $\lim_{n \to \infty} x_n = X$, show that $\frac{1}{\ln n}\sum_{j=1}^n\frac{x_j}{j}$ converges to $X$.
(2) For a sequence $\{x_n\}$ with convergent arithmetic means, $\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^n x_j = X$, show that $\frac{1}{\ln n}\sum_{j=1}^n\frac{x_j}{j}$ converges to $X$.
Attempt:
For (1) I used the Cesaro Stolz theorem:
$$\lim_{n \to \infty}\frac{1}{\ln n}\sum_{j=1}^n\frac{x_j}{j} = \lim_{n \to \infty}\frac{\frac{x_{n+1}}{n+1}}{\ln(n+1)-\ln(n)} = \lim_{n \to \infty}\frac{x_{n+1}}{\ln(1+1/n)^{n+1}} = \frac{X}{\ln(e)} = X$$
For (2) I know if all $x_n > 0$ then $\frac{1}{\ln n}\sum_{j=1}^n\frac{x_j}{j} < \frac{n}{\ln n}\frac{1}{n}\sum_{j=1}^n x_j$ but this does not seem to help.  Also it is not given that all $x_n >0$
Thank you for any assistance.
 A: Replacing $x_n$ by $x_n-X$ if necessary, we can assume that $X=0$. Let $s_n:=\sum_{i=1}^nx_i$. Then there exists a sequence $\left(\delta_n\right)_{n\geqslant 1}$ converging to $0$ such that $x_n=s_n-s_{n-1}=n\delta_n-\left(n-1\right)\delta_{n-1}$ hence 
$$
\frac 1{\ln n}\sum_{j=1}^n\frac{x_j}j=\frac 1{\ln n}\sum_{j=1}^n\frac{j\delta_j-\left(j-1\right)\delta_{j-1}}j=\frac 1{\ln n}\sum_{j=1}^n\left(\delta_j-\delta_{j-1}\right)+\frac 1{\ln n}\sum_{j=1}^n\frac{\varepsilon_{j-1}}j.
$$
The first sum in the last expression is telescopic and goes to $0$ as $n$ goes to infinity; for the second one, fix an integer $N$ and observe that for all $n\geqslant N$, 
$$
\left\lvert \frac 1{\ln n}\sum_{j=1}^n\frac{\varepsilon_{j-1}}j\right\rvert \leqslant \frac N{\ln n}+\frac 1{\ln n}\sum_{j=1}^n\frac{1}j\sup_{i\geqslant N}\left\lvert \delta_i\right\rvert.
$$
Since the quantity $\frac 1{\ln n}\sum_{j=1}^n\frac{1}j$ can be bounded independently on $n$ (say by $c$, we get
$$
\limsup_{n\to +\infty}\left\lvert \frac 1{\ln n}\sum_{j=1}^n\frac{\varepsilon_{j-1}}j\right\rvert \leqslant c\sup_{i\geqslant N}\left\lvert \delta_i\right\rvert
$$
and since $N$ is arbitrary, we can conclude.
A: Summation by parts gives:
\begin{align}
\sum_{j=1}^n \frac{x_j}j &= \frac1n \sum_{j=1}^n x_j - \sum_{j=0}^{n-1} \left(\sum_{i=1}^jx_i\right)\left(\frac1{j+1} - \frac1j\right) \\
&= \frac1n \sum_{j=1}^n x_j - \sum_{j=0}^{n-1} \left(\frac1j\sum_{i=1}^jx_i\right)\left(\frac{j}{j+1} - 1\right)\\
&= \sum_{j=0}^{n-1} \frac1{j+1}\left(\frac1j\sum_{i=1}^jx_i\right) + \frac1n \sum_{j=1}^n x_j \\
\end{align}
The assumption is $\lim_{n\to\infty} \frac1n \sum_{i=1}^n x_i = X$ so also $\lim_{n\to\infty} \frac1{n-1} \sum_{i=1}^{n-1} x_i = X$. The first part gives
$$X = \lim_{n\to\infty} \frac1{\ln n}\sum_{j=1}^n \frac{\frac1{j-1} \sum_{i=1}^{j-1} x_i}{j} = \lim_{n\to\infty} \frac1{\ln n}\sum_{j=0}^{n-1} \frac{\frac1{j} \sum_{i=1}^{j} x_i}{j+1} = \lim_{n\to\infty} \frac1{\ln n}\left[\sum_{j=1}^n \frac{x_j}j - \frac1n \sum_{j=1}^n x_j\right]$$
so
$$\lim_{n\to\infty} \frac1{\ln n} \sum_{j=1}^n \frac{x_j}j = X + \lim_{n\to\infty} \frac1{\ln n}\left(\frac1n \sum_{j=1}^n x_j\right) = X$$
