Find a simple sequence that is asymptotically equivalent to $u_n = \sum_{k=n+1}^{+\infty} \frac{\ln{k}}{k^2}$ 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.
 A: 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 $n\to +\infty$.
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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)}
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\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}
