Let $S$ be the sum of the series $\displaystyle\sum_{n=2}^{\infty} \frac {1}{n\ln (n)^2}$. We show that $\displaystyle\int_{n=2}^{c} \frac {1}{x\ln (x)^2}dx=\frac{1}{\ln2}-\frac{1}{\ln c}$ so the series is convergent. Now we want a lower and an upper bound for $S$. We know that $$S_n+\int_{n+1}^{\infty}\frac {1}{x\ln (x)^2}dx \le S \le S_n+\int_{n}^{\infty}\frac {1}{x\ln (x)^2}dx$$ My question is how to find the value of $n$ that gives the lowest $S_n+\int_{n}^{\infty}\frac {1}{x\ln (x)^2}dx$ and the biggest $S_n+\int_{n+1}^{\infty}\frac {1}{x\ln (x)^2}dx$
Thank you for your help ?