If $\lim\limits_{n\to \infty}a_n=a$, then what is the value of $\lim\limits_{n\to \infty}\frac{1}{\ln(n)}\sum_{r=1}^{n}\frac{a_r}{r}$?

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\bracks{{1 \over \ln\pars{n}}\sum_{r = 1}^{n}{a_{r} \over r}} & = \lim_{n \to \infty}\bracks{{1 \over \ln\pars{n + 1} - \ln\pars{n}} \pars{\sum_{r = 1}^{n + 1}{a_{r} \over r} - \sum_{r = 1}^{n}{a_{r} \over r}}} \\[5mm] & = \lim_{n \to \infty}{a_{n + 1}/\pars{n + 1} \over \ln\pars{1 + 1/n}} \qquad\pars{~Stolz-Ces\grave{a}ro Theorem~} \\[5mm] & = \lim_{n \to \infty}\bracks{% {1/n \over \ln\pars{1 + 1/n}}\,{n \over n + 1}\,a_{n + 1}} = \bbx{\ds{a}} \end{align}

See this link about Stolz-Cesàro theorem Theorem.

We can easily show that

$$\lim_{n\to \infty}\frac{1}{\log(n)}\sum_{r=1}^n\frac{a}{r}=a$$

by using the bounds

$$\frac{1}{\log(n)}\int_1^n \frac{a}x\,dx\le \frac{1}{\log(n)}\sum_{r=1}^n\frac{a}{r}\le \frac{1}{\log(n)}\left(1+\int_{1}^n\frac{a}{x}\,dx\right)$$

and applying the squeeze theorem.

---

Note that for any $\epsilon>0$ there exists a number $N$ such that $|a_r-a|<\epsilon/2$ whenever $n>N$. Then, we have$$\begin{align}\left|\frac{1}{\log(n)}\sum_{r=1}^n\frac{a_r-a}{r}\right|&=\left|\frac{1}{\log(n)}\sum_{r=1}^N\frac{a_r-a}{r}+\frac{1}{\log(n)}\sum_{r=N+1}^n\frac{a_r-a}{r}\right|\\\\&\le \frac{1}{\log(n)}\sum_{r=1}^N\frac{|a_r-a|}{r}+\frac{\epsilon/2}{\log(n)}\sum_{r=N+1}^n \frac{1}{r}\tag1\end{align}$$For fixed $N$, the first sum on the right-hand side goes to $0$ as $n\to \infty$.

For the second term, we know that $\sum_{r=N+1}^n \frac{1}{r}\le \int_{N}^n \frac1x\,dx=\log(n)-\log(N)$. Therefore, the entire right-hand side of $(1)$ can be made smaller than any pre-assigned number $\epsilon$ by selecting $N$ large enough.

Therefore, we find that

$$\lim_{n\to \infty}\frac{1}{\log(n)}\sum_{r=1}^n\frac{a_r}{r}=\lim_{n\to \infty}\frac{1}{\log(n)}\sum_{r=1}^n\frac{a}{r}=a$$

and we are done!