My question is: simplify the following $$S=\mathcal{O}\left(\sum_{m=2}^{n}\frac{\log m}{m^2}\right)$$ where $\mathcal{O}$ denotes Big O notation.
Wolfram alpha shows: https://www.wolframalpha.com/input?i=sum+%28log+m%29%2Fm%5E2%2C+m%3D1+to+n. But it is not mentioned there that what does $\zeta^(1,0)(2,n+1)$ and $A$ mean.
I tried using $$\sum_{m=2}^{n}\frac{\log m}{m^2}\leq \int_{2}^{n} \frac{\log x}{x^2}dx$$
And hence we have $$ \sum_{m=2}^{n}\frac{\log m}{m^2} \leq \frac{1}{2}\left(1+\log 2-\frac{2(1+\log n)}{n}\right)$$
I need a tight upper bound. Please help in this regard.
As Gerry Myerson mentioned, a tight upper bound is the limit as $n\to\infty$ of the partial sums. If you are interested in an even tighter bound, you can look at the sequence of tails $\sum_{m=N}^\infty\frac{\log m}{m^2}$ since $\sum_{m=1}^{N-1}\frac{\log m}{m^2}=\sum_{m=1}^{\infty}\frac{\log m}{m^2} - \sum_{m=N}^{\infty}\frac{\log m}{m^2}$. We have the estimate $\sum_{m=N}^{\infty}\frac{\log m}{m^2} \ge \int_N^\infty \frac{\log x}{x^2}\,dx=\frac{1}{N}+\frac{\log N}{N}$. Thus an upper bound for the $N-1$-th partial sum is $\left(\sum_{m=1}^{\infty}\frac{\log m}{m^2}\right)-\frac{1}{N}-\frac{\log N}{N}=-\zeta'(2)-\frac{1}{N}-\frac{\log N}{N}$.