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

Solution 1:

$$\sum_{k=1}^\infty\frac{1}{k^2+3k+1}\geq\sum_{k=1}^\infty\frac{1}{k^2+3k+2}=\sum_{k=1}^\infty\left(\frac{1}{k+1}-\frac{1}{k+2}\right)=$$ $$\lim_{n\to\infty}\;\;\sum_{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)$$ 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 $$\sum_{k=1}^\infty\frac{1}{k^2+3k+1}\geq\frac{1}{2}$$

Solution 2:

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*}