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
 A: 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*}
A: An ugly approach: evaluate the partial sum directly (and notice each term is positive).
$$
\sum_{k=1}^\infty \frac1{k^2+3k+1} > \sum_{k=1}^{20}\frac1{k^2+3k+1} = 0.5007647\dots $$
A: 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 $0.5012485$.)
A: $$\sum_{k=1}^\infty\frac{1}{k^2+3k+1}\geq\sum_{k=1}^\infty\frac{1}{k^2+3k+2}=\sum_{k=1}^\infty\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=$$
$$\lim_{n\to\infty}\;\;\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)$$
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
$$\sum_{k=1}^\infty\frac{1}{k^2+3k+1}\geq\frac{1}{2}$$
