# Tight upper bound for $\sum_{m=2}^n \frac{ \log m}{m^2}$.

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 $$\sum_{m=2}^{n}\frac{\log m}{m^2} \leq \frac{1}{2}\left(1+\log 2-\frac{2(1+\log n)}{n}\right)$$

• The infinite sum converges, so for large $n$ a tight upper bound is the sum of the infinite series. Commented Mar 10, 2022 at 11:21
• @GerryMyerson Thanks. Can we write $S=\mathcal{O}(\log n/n)$?
– user1034301
Commented Mar 10, 2022 at 11:34
• I take it you mean $(\log n)/n$, and not $\log(n/n)$. But $(\log n)/n\to0$ as $n\to\infty$, while $S$ has a finite nonzero limit. Commented Mar 10, 2022 at 11:54
• @GerryMyerson So what is the big O bound for this sum?
– user1034301
Commented Mar 10, 2022 at 12:07
• It is $O(1)$, as pointed out by Gerry. Commented Mar 10, 2022 at 18:49

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 $$\left(\sum_{m=1}^{\infty}\frac{\log m}{m^2}\right)-\frac{1}{N}-\frac{\log N}{N}=-\zeta'(2)-\frac{1}{N}-\frac{\log N}{N}$$.