I found a formula for $\pi$ in this paper. However, I could not find any proof of this formula, and I don't know how to approach to it. Is there good explanation for it?
$$ \pi + 3 = \sum_{n=1}^\infty \frac{n2^nn!^2}{(2n)!} $$
Proving a formula for $\pi$
pisequences-and-series
Related Solutions
I know this general result: $$\displaystyle \sum_{k=1}^{\infty}\frac{\cos\bigl({k\pi-\sqrt{k^2\pi^2-a^2}\,}\bigr)}{k^2}=\frac{\pi^2}{12}\,\bigl({-\cosh(a)+\frac{3}{a}\sinh(a)}\bigr) $$
and you can see this paper:On some strange summation formulas by R. William Gosper 1993 can download by:http://projecteuclid.org/euclid.ijm/1255987146
1. The $\operatorname{arctan}$ formulas.
Consider what it means for a complex number $u$ to be a logarithm of $z$: essentially, that $e^u = z$.
Since $e^u = e^{\Re u + i \Im u} = e^{\Re u} e^{i\Im u} = r e^{i \theta}$ you see that pretty much by definition, $\Im \ln(z) = \Im u = \theta$ is the angular component of $z$ in the complex plane.
Now, take the point $z = 1-iy$, is it an elementary trigonometric exercise to check that the angle forme between the real axis and $z$ is $\theta$ such that $\tan\theta = -y$. By the above remark this means exactly that $\Im \ln(1-iy) = \arctan(-y)$.
Since $\arctan$ is an odd function, this means that $-\Im \ln(1-iy) = \arctan(y)$.
The very same argument works for $z = 1 - (1+i)/a$. Indeed, this time we have
$$\begin{align*} \tan \theta & = \frac{\Im z}{\Re z} \\ & = \frac{-1/a}{1 - 1/a} \\ & = \frac{-1}{a - 1} \end{align*} $$ And therefore, $-\Im\ln(1 - (1+i)/a) = \arctan(1/(a-1))$.
2. The summation formula
You have it right, except that $(1+i)^3 = 2i(1+i) = -2 + 2i$. And not $2 - 2i$ as you wrote, which was the source of this erroneous sign.
Well done! For additional fun formulas and the rationale behind them (in French), you can have a look at this pdf from the olden days.
Best Answer
Here's part of an exercise from the book Pi and the AGM by Borwein and Borwein.
Prove that $$\frac{2x\sin^{-1}x}{\sqrt{1-x^2}}=\sum_{m=1}^\infty \frac{m!^2(2x)^{2m}}{m(2m)!}.\tag{1}$$ Hint: show that $f=(\sin^{-1}x)/\sqrt{1-x^2}$ satisfies $(1-x^2)f'=1+xf$.
Granted $(1)$, differentiating and multiplying by $x$ gives a formula for $$\sum_{m=1}^\infty \frac{m!^2(2x)^{2m}}{(2m)!}.$$ Doing it again, gives a formula for $$\sum_{m=1}^\infty m\frac{m!^2(2x)^{2m}}{(2m)!}.$$ Finally set $x=1$.