Proof about an infinite sum: $\sum_{k=1}^\infty \frac1{k^2+3k+1} \ge \frac12$

Hello I have a pretty elementary question but I am a bit confused.

I am trying to prove that $$\sum_{k=1}^\infty \frac1{k^2+3k+1} \ge \frac12$$

thanks, Thrasyvoulos

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$$\sum_{k=1}^\infty\frac{1}{k^2+3k+1}\geq\sum_{k=1}^\infty\frac{1}{k^2+3k+2}=\sum_{k=1}^\infty\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=$$ $$\lim_{n\to\infty}\;\;\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)$$ Note that $$\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{2}-\frac{1}{n+2}$$ so $$\lim_{n\to\infty}\;\;\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=\lim_{n\to \infty}\left(\frac{1}{2}-\frac{1}{n+2}\right)=\frac{1}{2}$$ Thus, we have shown $$\sum_{k=1}^\infty\frac{1}{k^2+3k+1}\geq\frac{1}{2}$$

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Beautiful clear proof. – orlp Nov 23 '11 at 14:25
Thank you for the kind words. I actually tried applying partial fraction decomposition to $k^2+3k+1$ first, saw that nothing nice was going to happen, then realized that $k^2+3k+2$ would telescope correctly and provide an approximation from the right direction. – Zev Chonoles Nov 23 '11 at 16:06
That's really elegant. – Chris Cudmore Nov 23 '11 at 17:32

We have $$\sum_{k\geq 1}\frac 1{k^2+3k+1}\geq \sum_{k\geq 1}\frac 1{k^2+4k}=\frac 14\sum_{k\geq 1}\frac {k+4-k}{k^2+4k}=\frac 14\sum_{k\geq 1}\left(\frac 1k-\frac 1{k+4}\right),$$ hence \begin{align*} \sum_{k\geq 1}\frac 1{k^2+3k+1} &\geq \frac 14\lim_{N\to\infty}\left(\sum_{k=1}^N\frac 1k-\sum_{k=1}^N\frac 1{k+4} \right)\\ &=\frac 14\left(\sum_{j=1}^N\frac 1j-\sum_{j=5}^{N+4}\frac 1j \right)\\ &=\frac 14\left(1+\frac 12+\frac 13+\frac 14-\frac 1{N+1}-\frac 1{N+2}-\frac 1{N+3}-\frac 1{N+4}\right)\\ &=\frac 14\left(\frac 32+\frac 7{12}\right)\\ &=\frac{6\cdot 3+7}{48}\\ &=\frac{25}{48}\geq\frac 12. \end{align*}

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An ugly approach: evaluate the partial sum directly (and notice each term is positive).

$$\sum_{k=1}^\infty \frac1{k^2+3k+1} > \sum_{k=1}^{20}\frac1{k^2+3k+1} = 0.5007647\dots$$

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Here's a "sledgehammer" approach that uses the fact that $\sum_{k=1}^\infty {1\over k^2}={\pi^2\over6}$:

\eqalign{ \sum_{k=1}^\infty {1\over k^2+3k +1}&= \sum_{k=1}^7 {1\over k^2+3k+1 } + \sum_{k=8}^\infty {1\over k^2+3k +1}\cr &\ge \sum_{k=1}^7 {1\over k^2+3k +1} + \sum_{k=8}^\infty {1\over 2 k^2 }\cr &= \sum_{k=1}^7 {1\over k^2+3k +1} + {1\over2}({\pi^2\over 6} -\sum_{k=1}^7 {1\over k^2})\cr &\ge 1/2.}

(Assuming I did the arithmetic correctly in the last step (the expression on the left of the inequality is approximately $0.5012485$.)

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