# 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?

• In those sums you’ve written out, everything cancels except finitely many terms. For example, in the first sum, the very next term inside the $\cdots$ is $\frac 1 9$ which cancels with the $\frac 19$ in the second sum. – User8128 Dec 8 '18 at 15:36

## 3 Answers

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?

Check that $$\left(\frac13+\frac15+\frac17+\frac19+\frac1{11}+\cdots+\frac1{2N+1}\right)-$$ $$-\left(\frac19+\frac1{11}+\cdots+\frac1{2N+1}+\frac1{2N+3}+\frac1{2N+5}+\frac1{2N+7}\right)=$$ $$=\frac13+\frac15+\frac17-\frac1{2N+3}-\frac1{2N+5}-\frac1{2N+7}.$$

And if you take the limit as $$N\to\infty$$ this becomes just $$\frac13+\frac15+\frac17.$$

After given

$$\sum_{n=1}^{\infty} \dfrac{1}{(2n+1)(2n+7)}=\frac1{36}\sum_{n=1}^{\infty} \dfrac{1}{(\frac{n}{3}+\frac{1}{6})(\frac{n}{3}+\frac{7}{6})}$$

set function

$$f(x)=\frac1{36}\sum_{n=1}^{\infty} \dfrac{x^{\frac{n}{3}+\frac{7}{6}}}{(\frac{n}{3}+\frac{1}{6})(\frac{n}{3}+\frac{7}{6})}$$

then take the second derivative of function $$f(x)$$, which is

$$f''(x)=\frac1{36}\sum_{n=1}^{\infty} x^{\frac{n}{3}-\frac{5}{6}}$$

and it is easy to find this series is equal to

$$f''(x)=\frac{1}{36x^{1/2}(1-x^{1/3})}$$

also, notice that you also get the boundary $$f'(0)=0$$ and $$f(0)=0$$, then you can do the integral twice to find the original $$f(x)$$, which is

$$f(x)=\frac{x^{1/6}}{6}(1+\frac{1}{3}x^{1/3}+\frac{1}{5}x^{2/3}-\frac{6}{7}x)+\frac{1-x}{12}\ln\left(\frac{1+x^{1/6}}{1-x^{1/6}}\right)$$

and take the limitation for $$x\to1$$ which is the result

$$\sum_{n=1}^{\infty} \dfrac{1}{4n^2+16n+7}= \lim_{x \to 1} f(x)=\frac{71}{630}$$

Actually, the general function method is much more complicated than fraction splitting.