<s>[INST] 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 $n \geq 1$:
$$\sum\limits_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{13}{20} .$$

 [/INST] 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.


*

*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
$$\sum_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{1}{5} + \int_1^{\infty} \frac{1}{k^2+3k+1} dk.$$

*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
$$\sum_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{1}{5} + \int_1^{\infty} \frac{1}{k^2+2k+1} dk = \frac{1}{5} + \frac{1}{2} = \frac{14}{20}.$$

*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.
</s>