TITLE: If $\lim\limits_{n\to \infty}a_n=a$, then what is the value of $\lim\limits_{n\to \infty}\frac{1}{\ln(n)}\sum_{r=1}^{n}\frac{a_r}{r}$?
QUESTION [5 upvotes]: For a given sequence $a_1,a_2,\ldots,a_n$ if $\lim\limits_{n\to \infty}a_n=a$, then $\lim\limits_{n\to \infty}\dfrac{1}{\ln(n)}\sum\limits_{r=1}^{n}\dfrac{a_r}{r}$ is:
(A)zero $\hspace{150pt}$ (B)a
(C)$\dfrac{a}{2}$ $\hspace{157pt}$ (D)None of these

My approach: I'm trying to bring this limit into this form:  $\lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{r=1}^{n}f\left(\dfrac{r}{n}\right)$ as, $$\lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{r=1}^{n}f\left(\dfrac{r}{n}\right)=\int_{0}^{1}f(x)\ dx$$
so here it is what I'm doing:
\begin{align*}
\lim\limits_{n\to \infty}\dfrac{1}{\ln(n)}\sum\limits_{r=1}^{n}\dfrac{a_r}{r}
&=\lim\limits_{n\to \infty}\dfrac{1}{n}\cdot\dfrac{n}{\ln(n)}\sum\limits_{r=1}^{n}\dfrac{a_r}{r}\\
&=\lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{r=1}^{n}\dfrac{a_r}{r/n}\cdot\dfrac{1}{\ln(n)}
\end{align*}
but I can't make the limit in the above form, is there any way of doing it?

REPLY [6 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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.