Show that: $$\sum_{k=2}^{n}\dfrac{\ln{k}}{k^2}\approx \ln{n}\cdot\left(\zeta_{n}{(2)}-\dfrac{\pi^2}{6}\right)+C,n\to\infty$$ where $$\zeta_{n}{(k)}=\sum_{j=1}^{n}\dfrac{1}{j^k}$$and $C$ is real constant.
I know this $$\zeta{(x)}=\sum_{n=1}^{\infty}\dfrac{1}{n^x}$$ $$\Longrightarrow \zeta'{(x)}=-\sum_{n=1}^{\infty}\dfrac{\ln{n}}{n^x}$$ let $x=2$,then we have $$\sum_{n=1}^{\infty}\dfrac{\ln{n}}{n^2}=\sum_{n=2}^{\infty}\dfrac{\ln{n}}{n^2}=-\zeta'{(2)}$$ But for this approximation,I can't prove it.
Thank you very much
Rewriting the sums a bit, you're asking how to show that
$$ \sum_{k=2}^{\infty} \frac{\ln k}{k^2} - \sum_{k=n+1}^{\infty} \frac{\ln k}{k^2} \approx -\ln n \sum_{k=n+1}^{\infty} \frac{1}{k^2} + C. $$
Looking at this one might guess that we'll probably have $C = \sum_{k=2}^{\infty} \frac{\ln k}{k^2}$, so we would just need to prove that
$$ \sum_{k=n+1}^{\infty} \frac{\ln k}{k^2} \approx \ln n \sum_{k=n+1}^{\infty} \frac{1}{k^2}. \tag{1} $$
The argument you would use depends on what exactly you mean by $\approx$. For example, to leading order
$$ \sum_{k=n+1}^{\infty} \frac{\ln k}{k^2} \approx \int_n^\infty \frac{\ln x}{x^2}\,dx = \frac{\ln n + 1}{n} \approx \frac{\ln n}{n} $$
and
$$ \sum_{k=n+1}^{\infty} \frac{1}{k^2} \approx \int_n^\infty \frac{dx}{x^2} = \frac{1}{n}, $$
so we have a proof of $(1)$ in this sense.