Equivalent of a remainder I want to find en equivalent of the rest of the convergent series for $n \in \mathbb{N}^{*}$
$$
R_n=\sum_{k=n+1}\frac{1}{k^2\ln\left(k\right)}
$$
I used that $\displaystyle x \mapsto \frac{1}{x^2\ln\left(x\right)}$ is positive, decreasing and tending to $0$ for $x \in \left[2,+\infty\right[$.
Hence
$$
\int_{k}^{k+1}\frac{\text{d}x}{x^2\ln\left(x\right)} \leq \frac{1}{k^2\ln\left(k\right)} \leq \int_{k-1}^{k}\frac{\text{d}x}{x^2\ln\left(x\right)}
$$
By chasles relation and summing to $N \geq n+1$
$$
\int_{n+1}^{N}\frac{\text{d}x}{x^2\ln\left(x\right)} \leq \sum_{k=n+1}^{N+1}\frac{1}{k^2\ln\left(k\right)} \leq \int_{n}^{N}\frac{\text{d}x}{x^2\ln\left(x\right)}
$$
But i'm idiotically stuck here, how can I compute this easily ? 
Do I need to find an equivalent of $\displaystyle x \mapsto \frac{1}{x^2\ln\left(x\right)}$ in $+\infty$ and integrate the comparison ?
I think that those integrals are so-called Bertrand's integrals

Since $\log n$ is approximately constant on short intervals and $\sum_{k>n}\frac{1}{k^2}$ behaves like $\frac{C}{n}$, it is reasonable to expect that $R_n \sim \frac{C}{n\log n}$. And by the Stolz-Cesàro theorem we have
$$ \lim_{n\to +\infty}\frac{R_n}{\frac{1}{n\log n}}=\lim_{n\to +\infty}\frac{R_n-R_{n+1}}{\frac{1}{n\log n}-\frac{1}{(n+1)\log(n+1)}}=\lim_{n\to +\infty}\frac{\frac{1}{(n+1)^2\log(n+1)}}{\frac{1}{n\log n}-\frac{1}{(n+1)\log(n+1)}}=\color{red}{1} $$
proving our conjecture with $C=1$.
