Evaluate $\lim_{n\to\infty}\sum_{k=1}^n \frac{\log(k)}{nk}$ 
Evaluate
$$S=\lim_{n\to\infty}\sum_{k=1}^n \frac{\log(k)}{nk}$$

My first thought was Riemann sum, but I don't have $\frac{k}{n}$, only $k$. Since I have no idea how to evaluate this, I tried the following. First, I tried to find a lower bound for $S$.
$$\sum_1^n\frac{\log(k)}{k}>\sum_1^n\frac{\log(k)}{n}=\frac{1}{n}\log\left(\frac{n^2+n}{2}\right)$$
Thus,
$$\frac{1}{n}\sum_{k=1}^n\frac{\log(k)}{k}>\frac{1}{n^2}\log\left(\frac{n^2+n}{2}\right)$$
Since
$$\lim_{n\to\infty}\frac{1}{n^2}\log\left(\frac{n^2+n}{2}\right)=\lim_{n\to\infty}\frac{1}{2n}\frac{2}{n^2+n}\left(n+\frac{1}{2}\right)=0$$
$S>\ge 0$, if exists at all.
For the upper bound, since
$$\sum_1^n\frac{\log(k)}{k}<\sum_1^n\frac{k}{k}=n$$
Thus,
$$\frac{1}{n}\sum_{k=1}^n\frac{\log(k)}{k}<\frac{1}{n}n=1$$
So $0\le S \le 1$. But I don't know how to proceed from here. I think I am overlooking something obvious. Any hint (preferably not full solution) is appreciated.
 A: We could also use the fact that if $a_n \to L,$ then $(a_1 + \cdots +a_n)/n \to L.$ In your problem we have $a_n = (\ln n)/n \to 0.$ Therefore
$$\frac{\sum_{k=1}^{n}(\ln k)/k}{n} \to 0$$
and we're done.
A: Observe
\begin{align}
\left|\frac{1}{n}\sum^n_{k=1}\frac{\log k}{k}\right| \leq \frac{1}{n}+\frac{1}{n}\int^{n+1}_1 \frac{\log x}{x}\ dx \leq \frac{1}{n}[1+\log^2(n+1)]\rightarrow 0
\end{align}
as $n\rightarrow \infty$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

By means of the $\ds{Stoltz\!-\!Ces\grave{a}ro\ Theorem}$:

\begin{align}
S & \equiv \lim_{n \to \infty}\sum_{k = 1}^{n}{\ln\pars{k} \over nk} =
\lim_{n \to \infty}{\sum_{k = 1}^{n + 1}\ln\pars{k}/k -
\sum_{k = 1}^{n}\ln\pars{k}/k \over \pars{n + 1} - n} =
\lim_{n \to \infty}{\ln\pars{n + 1} \over n + 1}
\\[5mm] & =
\lim_{n \to \infty}{\ln\pars{\bracks{n + 1} + 1} - \ln\pars{n + 1} \over
\bracks{\pars{n + 1} + 1} - \pars{n + 1}}=
\lim_{n \to \infty}\ln\pars{n + 2 \over n + 1} =
\lim_{n \to \infty}\ln\pars{1 + {1 \over n + 1}} =
\bbx{\ds{0}}
\end{align}
A: Using Abel's summation we can see that $$S=\sum_{k=1}^{n}\frac{\log\left(k\right)}{k}=H_{n}\log\left(n\right)-\int_{1}^{n}\frac{H_{t}}{t}dt
 $$ where $H_{n}
 $ is the $n$-th harmonic number and since $H_{n}=\log\left(n\right)+O\left(1\right)
 $ we have $$S=\log^{2}\left(n\right)-\int_{1}^{n}\frac{\log\left(t\right)}{t}dt+O\left(\log\left(n\right)\right)
 $$ $$=\frac{\log^{2}\left(n\right)}{2}+O\left(\log\left(n\right)\right)
 $$ so $$\lim_{n\rightarrow\infty}\frac{S}{n}=\color{red}{0}.$$
