TITLE: Prove that $\sum\limits_{k=1}^n \frac{1}{k^2+3k+1}$ is bounded above by $\frac{13}{20}$
QUESTION [10 upvotes]: I want ask a question about a sum. The exercise is as follows:

Prove the following inequality for every $n \geq 1$:
$$\sum\limits_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{13}{20} .$$

REPLY [2 votes]: This time i will appeal to the famous result of the Basel problem and get that:
$$\sum\limits_{k=1}^\infty \frac{1}{k^2+3k+1} \leq \sum\limits_{k=1}^\infty \frac1{(k+1)^2} = \frac{{\pi}^2}{6}-1\\$$
But $$\frac{{\pi}^2}{6}-1 \leq \frac{13}{20}$$
The proof is complete.