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 $n \geq 1$:

$$\sum\limits_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{13}{20} .$$

Since $\frac{1}{k^2+3k+1}$ is monotone decreasing for $k\geq 0$, we have

$$\begin{align*}\sum_{k=1}^n \frac{1}{k^2+3k+1} &\leq \frac{1}{5} + \frac{1}{11} + \int_2^\infty \frac{1}{k^2+3k+1} dk\\ &< \frac{1}{5} + \frac{1}{11} + \int_2^\infty \frac{1}{k^2+2k+1} dk\\ &= \frac{1}{5}+ \frac{1}{11} + \frac{-1}{k+1}\Big\vert_2^\infty\\ &= \frac{1}{5} + \frac{1}{11} + \frac{1}{3}\\ &< \frac{13}{20}. \end{align*} $$

EDIT: I didn't realize this was tagged homework; I now feel a little guilty giving such an explicit solution. Here are the steps I took to get at this answer, which might be useful for solving similar problems.

1. I remembered that monotonic series can be bounded by integrals, by thinking of the series as a right Riemann sum. This suggests I try the bound $$\sum_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{1}{5} + \int_1^{\infty} \frac{1}{k^2+3k+1} dk.$$

2. That integral on the right looks mighty unpleasant; the denominator doesn't factor so the antiderivative will have logs and arctans galore. But I can bound the integral by the much nicer perfect square $$\sum_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{1}{5} + \int_1^{\infty} \frac{1}{k^2+2k+1} dk = \frac{1}{5} + \frac{1}{2} = \frac{14}{20}.$$

3. Ack! The bound is barely not tight enough. Pulling out more terms from the sum should improve it, so I try $$\sum_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{1}{5} + \frac{1}{11} + \int_2^{\infty} \frac{1}{k^2+2k+1} dk,$$ which after working out the details turns out to work.

$$ \begin{align} \sum\limits_{k=1}^n \frac{1}{k^2+3k+1} & \leq \sum\limits_{k=1}^n \frac1{k(k+3)}\\ & = \sum\limits_{k=1}^n \frac13 \left(\frac1k - \frac1{k+3} \right)\\ & = \frac13 \left( 1 + \frac12 + \frac13 - \frac1{n+1} - \frac1{n+2} - \frac1{n+3} \right)\\ & \leq \frac13 \frac{11}{6}\\ & = \frac{11}{18} \end{align} $$

(I noticed it just now. It is the same as Zarrax's and David Mitra's comments)