Show $\sum\limits_{n=0}^{\infty}\frac{1}{n(n+1)+x} = 2 \pi$ for some $x>0$ Prove that the series $\sum\limits_{n=0}^{\infty}\frac{1}{n(n+1)+x}$ for $x&gt;0$ contains a value c such that $f(c)=2\pi$.


It seems like this is somehow related to the fact that 
$$1-1/3+1/5-...=\pi/4$$
but I can't figure out how to manipulate the proof for that equation to align with this problem. 

I know that I can use the Mean Value Theorem to show this since when $x$ approaches $0$, the sum is 1. I just need to show that a value for $x$ exists so that $f(x)&gt;2\pi$.
Let $$f(x) = \sum_0^\infty \frac1{n(n+1)+x}$$.
Then for all $x&gt;0$
$$
f(x) \leq \frac1x + \sum_1^\infty \frac1{n(n+1)+x}  \leq \frac1x + \sum_1^\infty \frac1{n(n+1)} $$

In particular,
$$
f(1) \leq 1+ \sum_1^\infty \frac1{n(n+1)} = 1+1 = 2 &lt; 2\pi
$$
On the other hand, 
$$f(\frac18) =8+ \sum_1^\infty \frac1{n(n+1)+\frac18} &gt;8 &gt; 2\pi
$$
then by the intermediate value theorem, ther is some $c$ in $[1,8]$ such that $f(c) = 2\pi$.

To be rigorous, you also actually need to prove that $f(x)$ is continuous, and that is somewhat harder.
A possible direct solution using Maple is as follows.

The series is computed as

$$f(x)=\frac{\displaystyle\pi\tan\left(\frac{\pi\sqrt{1-4x}}2\right)}{\sqrt{1-4x}}$$

Then we get the equation 

$$\frac{\displaystyle\pi\tan\left(\frac{\pi\sqrt{1-4c}}2\right)}{\sqrt{1-4c}}=2\pi$$

and the corresponding solution is $$ c = 0.187500$$.

Do you agree?