TITLE: Proof about an infinite sum: $\sum_{k=1}^\infty \frac1{k^2+3k+1} \ge \frac12$
QUESTION [11 upvotes]: 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

REPLY [6 votes]: An ugly approach: evaluate the partial sum directly (and notice each term is positive).
$$
\sum_{k=1}^\infty \frac1{k^2+3k+1} > \sum_{k=1}^{20}\frac1{k^2+3k+1} = 0.5007647\dots $$