Find a simple sequence that is asymptotically equivalent to $u_n = \sum_{k=n+1}^{+\infty} \frac{\ln{k}}{k^2}$

Question

Given that $$\int_x^{+\infty} \frac{\ln t}{t^2}\ dt = \frac{\ln x+1}{x} \sim \frac{\ln x}{x}$$

I believe that $$u_n= \sum_{k=n+1}^{+\infty} \frac{\ln{k}}{k^2}\sim\frac{\ln n}{n}$$ as $n \to +\infty$. Numerical evidence suggests that this is the case.

Using Riemann sums on the function $f(t) = \ln t\ /\ t^2$ on the interval $[1/n,1]$ I was able to show that $u_n =O(n/\ln{n})$ and $n/\ln{n} = O(u_n)$, but I couldn't show that the two sequences are equivalent.

Answer 1

Your conjecture is correct. Observe

$$\int_n^\infty\frac{\ln t}{t^2}\,dt < \sum_{k=n}^{\infty}\frac{\ln k}{k^2} < \int_{n-1}^\infty\frac{\ln t}{t^2}\,dt$$

and use the given result for the integrals.

Answer 2

In order to prove that $u_n=\sum_{k>n}\frac{\log k}{k^2}\sim\frac{\log n}{n}$ it is enough to notice that

$$ \int_{n}^{+\infty}\frac{\log(x)}{x^2}\,dx=\frac{1+\log n}{n}$$ and that $\frac{\log x}{x^2}$ is convex on $[2.301,+\infty)$. In particular, by the Hermite-Hadamard inequality $$ \int_{n}^{n+1}\frac{\log x}{x^2}\,dx \leq \frac{1}{2}\left(\frac{\log n}{n^2}+\frac{\log(n+1)}{(n+1)^2}\right) $$ holds for any $n\geq 3$ and $$ 0\leq u_n-\int_{n}^{+\infty}\frac{\log(x)}{x^2}\,dx\leq \frac{1}{2}\sum_{k\geq n}\left(\frac{\log k}{k^2}-\frac{\log(k+1)}{(k+1)^2}\right)=\frac{\log n}{2n^2}. $$ The same trick applies with greater generality: If $\varphi:\mathbb{R}^+\to\mathbb{R}^+$ is a convex function on $(a,+\infty)$, $\sum_{k\geq 1}\varphi(k)$ is finite and $\int_{n}^{+\infty}\varphi(x)\,dx=F(n)$, then $\sum_{k>n}\varphi(k)\sim F(n)$ as $n\to +\infty$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} u_{n} & \equiv \sum_{k\ =\ n + 1}^{\infty}{\ln\pars{k} \over k^{2}} = \sum_{k\ =\ 1}^{\infty}{\ln\pars{k + n} \over \pars{k + n}^{2}} = {1 \over n^{2}}\sum_{k\ =\ 1}^{\infty} {\ln\pars{n} + \ln\pars{k/n} \over \pars{k/n + 1}^{2}} \\[5mm] & = \bracks{% {1 \over n}\sum_{k = 1}^{\infty}{1 \over \pars{k/n + 1}^{2}}} {\ln\pars{n} \over n} + \bracks{% {1 \over n}\sum_{k = 1}^{\infty}{\ln\pars{k/n} \over \pars{k/n + 1}^{2}}}{1 \over n} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,\ \underbrace{\bracks{\int_{0}^{\infty}{\dd x \over \pars{x + 1}^{2}}}} _{\ds{=\ 1}}\ {\ln\pars{n} \over n} +\ \underbrace{\bracks{\int_{0}^{\infty}{\ln\pars{x} \over \pars{x + 1}^{2}} \,\dd x}}_{\ds{=\ 0}}\ {1 \over n} = \bbx{\ln\pars{n} \over n} \end{align}