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

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


$$
\begin{align}
\sum\limits_{k=1}^n \frac{1}{k^2+3k+1}  & \leq \sum\limits_{k=1}^n \frac1{k(k+3)}\\
& = \sum\limits_{k=1}^n \frac13 \left(\frac1k - \frac1{k+3} \right)\\
& = \frac13 \left( 1 + \frac12 + \frac13 - \frac1{n+1} - \frac1{n+2} - \frac1{n+3} \right)\\ & \leq \frac13 \frac{11}{6}\\
& = \frac{11}{18}
\end{align}
$$
(I noticed it just now. It is the same as Zarrax's and David Mitra's comments)
