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?
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