If $\lim\limits_{n\to \infty}a_n=a$, then what is the value of $\lim\limits_{n\to \infty}\frac{1}{\ln(n)}\sum_{r=1}^{n}\frac{a_r}{r}$? 
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?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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.

A: 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!
A: I would start like this.
If $\lim_{n\rightarrow\infty} a_n = a$, then for any $\epsilon > 0$ (pick an $\epsilon$ close to 0) there is some large enough $N$ so that every $a_n$ is within $\epsilon$ of $a$ when $n>N$.  That is, as $n$ gets large, you can just say that $a_n$ is approximately $a$, and so $a_n/n$ is approximately $a/n$.
Then, since you know the technique of integral approximation, use it to show that $\sum_1^n a_r/r$ is approximately $a \ln n $ plus a constant (or look up partial sum of harmonic series, I'm suggesting show the result in Dr. MV's answer) and take your limit from there.
I put in "plus a constant" because it can take care of the error from your choice of $\epsilon$ and the fact that the first $N$ terms of $a_n$ that might not be close to $a$ at all.
A: Hint: What if $a_n = 1$ for all $n?$
