sequences and series - What value does $sum_{n=1}^{infty} dfrac{1}{4n^2+16n+7}$ converge to?

What value does

$$\sum_{n=1}^{\infty} \dfrac{1}{4n^2+16n+7}$$
converge to?

Ok so I've tried changing the sum to:

$$\sum_{n=1}^{\infty} \dfrac{1}{6(2n+1)}-\dfrac{1}{6(2n+7)}$$

and then writting some values:

$$\frac16·(\frac13+\frac15+\frac17\dots+\frac1{2N+1})-\frac16·(\frac19+\frac1{11}+\frac1{13}\dots+\frac1{2N+7})$$

but I don't know what else I can do to finish it! Any hint or solution?

Answer

Hint: Let's look at the $100$th partial sum. It's good to get some concreteness.

$$\frac{1}{6}\left(\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{201}\right)-\frac{1}{6}\left(\frac{1}{9}+\cdots+\frac{1}{205}+\frac{1}{207}\right).$$

We have a bunch of terms that are repeated: $\frac{1}{9}+\cdots+\frac{1}{201}$ exists in each bracketed portion, so we can simply cancel all of them out to get

$$\frac{1}{6}\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}-\frac{1}{203}-\frac{1}{205}-\frac{1}{207}\right).$$

Can you see how to use this line of reasoning to get the answer?