# 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 $n \geq 1$:

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

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Related post: math.stackexchange.com/q/84899. (By the way, @Maria, if you posted the other question through a different account, then you could merge the two accounts by flagging the moderators. If that was posted by someone else, please ignore my comment. ;)) – Srivatsan Nov 24 '11 at 18:06
${1 \over k^2 + 3k + 1} < {1 \over k^2 + 3k}$. Now use partial fractions on ${1 \over k^2 + 3k}$. – Zarrax Nov 24 '11 at 18:14
Just to nitpick, that $i$ should really be $k$ :) – Alexei Averchenko Nov 24 '11 at 18:18
Using Zarrax' comment, and writing down the resulting telescoping series, you'll see the partial sums are bounded by 11/18. – David Mitra Nov 24 '11 at 18:19
A quick way is to notice that $$\frac{1}{k^2+3k+1}\leq \frac{1}{k^2+2k+1}=\frac{1}{(k+1)^2}.$$ Then extending the series to infinity, this is $$\sum_{k=1}^\infty \frac{1}{(k+1)^2}=\zeta(2)-1=\frac{\pi^2}{6}-1\approx 0.644934\leq 0.65 =\frac{13}{20}.$$ – Eric Naslund Nov 24 '11 at 18:19

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.

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\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)

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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.

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Just curious. You are saying $\frac1 4=\frac{\pi^2}{6}-1$ right? – E.O. May 31 '12 at 7:35
@E.O.: OK. Thanks. – user 1618033 May 31 '12 at 7:39