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 $x>0$ 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 $f(x)>2\pi$.
 A: Let $$f(x) = \sum_0^\infty \frac1{n(n+1)+x}$$.
Then for all $x>0$
$$
f(x) \leq \frac1x + \sum_1^\infty \frac1{n(n+1)+x}  \leq \frac1x + \sum_1^\infty \frac1{n(n+1)} $$
In particular,
$$
f(1) \leq 1+ \sum_1^\infty \frac1{n(n+1)} = 1+1 = 2 < 2\pi
$$
On the other hand, 
$$f(\frac18) =8+ \sum_1^\infty \frac1{n(n+1)+\frac18} >8 > 2\pi
$$
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: Starting from your initial idea of : $\quad \displaystyle \frac{\pi}{4}=1-\frac 13+\frac 15-...$
We can write :
$$2\pi=8\times\frac{\pi}4=8\sum\limits_{n=0}^{\infty}\frac {(-1)^n}{2n+1}=\sum\limits_{n=0}^{\infty}\left(\frac {8}{4n+1}-\frac {8}{4n+3}\right)\\\sum\limits_{n=0}^{\infty}\frac {16}{(4n+1)(4n+3)}=\sum\limits_{n=0}^{\infty}\frac {16}{16n^2+16n+3}=\sum\limits_{n=0}^{\infty}\frac 1{n(n+1)+\frac 3{16}}$$

Thus $f(\frac 3{16})=2\pi$
A: The sum is uniformly convergent on each interval $[\epsilon, 1]$ with $\epsilon > 0$ and thus converges to a continuous function there. Furthermore, it is clearly decreasing on $(0, 1]$ and unbounded as $x\to 0$. Since
\begin{align*}
f(1) &= \sum_{n=0}^\infty \frac{1}{n(n+1) + 1} \leq 1 + \sum_{n=1}^\infty \frac{1}{n(n+1)} = 2 < 2\pi,
\end{align*}
it follows that $f(x) = 2\pi$ for some $x\in (0, 1]$.
A: 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?
