Behaviour of the sum $\sum\limits_{k\ge1}\frac{\log(t+k)}{k^2}$ 
What can be said about the behaviour of the sum $\displaystyle\sum\limits_{k\ge1}\frac{\log(t+k)}{k^2}\quad\mbox{if}\quad t>0$.

is it $O\left(\log\left(t\right)\right)$ ?.
 A: Hint : For a fixed $t$, you will have in one hand
$$\frac{\log(t)}{k^2}\leq \frac{\log(t+k)}{k^2}$$
and in, another hand, you can show that for all $k\geq 1$ (for $t\geq 2$)
$$ \frac{\log(t+k)}{\sqrt{k}}\leq  \log(t+1) $$
so (for $t\geq 2$)
$$\log(t)\sum \frac{1}{k^2}\leq \sum  \frac{\log(t+k)}{k^2}\leq \log(t+1)\sum \frac{1}{k\sqrt{k}} $$
and since you know that $\sum \frac{1}{k^2}$ and $\sum \frac{1}{k\sqrt{k}}$ are convergent (and of course with additionally $\log(t+1)\sim \log (t)$)....
A: $$\sum_{k\geq 1}\frac{1}{k^2}\int_{1}^{k+t}\frac{dx}{x}=\frac{\pi^2}{6}\log(t+1)+\sum_{k\geq 1}\frac{1}{k^2}\int_{1}^{k}\frac{dx}{x+t}\leq \frac{\pi^2}{6}\log(t+1)+\sum_{k\geq 1}\frac{\log k}{k^2} $$
hence the LHS is $\frac{\pi^2}{6}\log t+O(1)$ as $t\to +\infty$.
A: For $t\ge1$,
$$
\begin{align}
\log(t+k)
&=\log(t)+\log\left(1+\frac kt\right)\\
&\le\log(t)+\log(1+k)
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k=1}^\infty\frac{\log(t+k)}{k^2}
&\le\sum_{k=1}^\infty\frac{\log(t)+\log(1+k)}{k^2}\\
&=\log(t)\frac{\pi^2}6+\sum_{k=1}^\infty\frac{\log(1+k)}{k^2}\\
&\le\log(t)\frac{\pi^2}6+\log(2)+\int_1^\infty\frac{\log(1+x)}{x^2}\,\mathrm{d}x\\[3pt]
&=\log(t)\frac{\pi^2}6+3\log(2)
\end{align}
$$
Thus, for $t\ge1$,
$$
\frac{\pi^2}6\log(1+t)\le\sum_{k=1}^\infty\frac{\log(t+k)}{k^2}\le\frac{\pi^2}6\log(t)+3\log(2)
$$
