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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.

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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.

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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 $n\rightarrow \infty$.

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  • $\begingroup$ Why do we integrate to $n+1$? Can I just do $\sum_1^n \frac{\log(k)}{k} \le \int_1^n \frac{\log(x)}{x}dx$? $\endgroup$ – 3x89g2 Oct 20 '16 at 20:32
  • $\begingroup$ @Misakov The bounds could be off, but I wanted just an upper bound. $\endgroup$ – Jacky Chong Oct 20 '16 at 20:37
  • $\begingroup$ I'm sorry but I still don't understand. The best I can do is $\sum_1^n a_n \le a_1 +\int_1^n$... $\endgroup$ – 3x89g2 Oct 20 '16 at 21:05
  • $\begingroup$ Okay. Then it follows $\sum^n_1 a_n \leq a_1 + \int^n_1 \leq a_1 + \int^{n+1}_1$. $\endgroup$ – Jacky Chong Oct 20 '16 at 21:06
  • $\begingroup$ Initially, when I wrote the inequality I actually use $\int^{n+1}_1$ but realize it wasn't good enough so I added $a_1$ but didn't bother to change $\int^{n+1}_1 $ to $\int^n_1$. $\endgroup$ – Jacky Chong Oct 20 '16 at 21:07
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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 $$\lim_{n\rightarrow\infty}\frac{S}{n}=\color{red}{0}.$$

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

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}

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