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.

  • 1
    $\begingroup$ The infinite sum converges, so for large $n$ a tight upper bound is the sum of the infinite series. $\endgroup$ Commented Mar 10, 2022 at 11:21
  • $\begingroup$ @GerryMyerson Thanks. Can we write $S=\mathcal{O}(\log n/n)$? $\endgroup$
    – user1034301
    Commented Mar 10, 2022 at 11:34
  • $\begingroup$ 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. $\endgroup$ Commented Mar 10, 2022 at 11:54
  • $\begingroup$ @GerryMyerson So what is the big O bound for this sum? $\endgroup$
    – user1034301
    Commented Mar 10, 2022 at 12:07
  • 2
    $\begingroup$ It is $O(1)$, as pointed out by Gerry. $\endgroup$
    – PhoemueX
    Commented Mar 10, 2022 at 18:49

1 Answer 1


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


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