Upper bound for $\sum_{i=2}^{n} 1/(i\log(i))$

Question

I tried to find the asymtotic upper bounds for these summations $$\sum_{i=2}^{n} 1/(i\log(i)) \text{ and } \sum_{i=2}^{n} 1/(i\log(i)\log\log(i)) .$$

My guess is that they might be bounded by $O(\log\log n)$ and $O(\log\log\log n)$ repectively, but I don't know how to show it. Could you please show me how to bound these?

All I know for now is that they both diverge by checking in Wolfram Alpha.

Answer 1

We have: $$\frac{d}{dx}\log(\log x)=\frac{1}{x\log x},\qquad \frac{d}{dx}\log(\log(\log(x)))=\frac{1}{x\log(x)\log\log(x)},\tag{1}$$ and both $\log(\log(x))$ and $\log(\log(\log x)))$ are concave increasing functions for $x\geq 3$, so the Hermite-Hadamard inequality proves your claim. It also proves that the hidden constant inside the $O(\ldots)$ notation is just $1$.