Sum $\sum_{n=1}^{\infty}\sum_{k=1}^{2n}\frac{(-1)^{n+1}(2\pi)^{2n}}{(2n-k+1)!}\left(\frac{1}{2\pi i}\right)^k$ I was working on this double sum, which I found somewhere on the Internet.
$$\sum_{n=1}^{\infty}\sum_{k=1}^{2n}\frac{(-1)^{n+1}(2\pi)^{2n}}{(2n-k+1)!}\left(\frac{1}{2\pi i}\right)^k$$
I did some manipulations, some simplifications and ended with this.
$$\dfrac{1}{2\pi i}\sum_{n=1}^{\infty}\left(\frac{\Gamma(2n,-2\pi i)}
{\Gamma(2n)}-1\right)$$
Using Wolfram|Alpha, I found that this sum converges really fast, so I tried truncating the sum, but it didn't work out as I wasn't able to get a closed form solution for the sum. I also tried approximating using integral, and ended with this
$$-\int_1^{\infty} \frac{\gamma(2n,-2\pi i)}{\Gamma(2n)}\ \! \mathrm dn $$
This can be further re-written using the power series representation of lower incomplete gamma function.
$$-\sum_{k=0}^{\infty}(-2\pi i)^k\int_1^{\infty}\frac{(-2\pi i)^{2n}}{(2n+k)!}\ \! \mathrm dn $$
This integral is really hard, I wasn't able to even start solving it. Please help me evaluating this sum or the integral. A truncated version of the sum would also work.
 A: The given sum is $$-\sum_{n=1}^\infty\sum_{k=1}^{2n}\frac{(2\pi i)^{2n-k}}{(2n-k+1)!}=-\frac{1}{2\pi i}\sum_{n=1}^\infty a_n,\quad a_n=\sum_{k=1}^{2n}\frac{(2\pi i)^k}{k!}$$ (after replacing $k$ by $2n-k+1$ in the inner sum). And $a_n$ converges to $e^{2\pi i}-1=0$ (decaying faster than any exponential in $n$) as $n\to\infty$. Now, for $|x|<1$ (exchanging the summations),
$$
S(x):=\sum_{n=1}^\infty a_n x^{2n}=\sum_{k=1}^\infty\frac{(2\pi i)^k}{k!}\sum_{n=\lceil k/2\rceil}^\infty x^{2n}\\=\sum_{k=1}^\infty\left(\frac{(2\pi i)^{2k}}{(2k)!}+\frac{(2\pi i)^{2k-1}}{(2k-1)!}\right)\frac{x^{2k}}{1-x^2}\\=\frac{\cos 2\pi x-1+ix\sin 2\pi x}{1-x^2}
$$
and $\sum\limits_{n=1}^\infty a_n=\lim\limits_{x\to 1^-}S(x)=-\pi i$ by Abel's theorem. Thus, the given sum equals $\boxed{1/2}$.
A: We have :
\begin{aligned}\sum_{n=1}^{+\infty}{\sum_{k=1}^{2n}{\frac{\left(-1\right)^{n+1}\left(2\pi\right)^{2n}}{\left(2n-k+1\right)!\left(2\pi\,\mathrm{i}\right)^{k}}}}&=-\frac{1}{2\pi\,\mathrm{i}}\sum_{n=1}^{+\infty}{\sum_{k=1}^{2n}{\frac{\left(2\pi\,\mathrm{i}\right)^{2n-k+1}}{\left(2n-k+1\right)!}}}\\ &=-\frac{1}{2\pi\,\mathrm{i}}\sum_{n=1}^{+\infty}{\sum_{k=1}^{2n}{\frac{\left(2\pi\,\mathrm{i}\right)^{k}}{k!}}}\\ &=\sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\left(2\pi\right)^{2n}}{\left(2n\right)!}\int_{0}^{1}{x^{2n}\,\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}}\ \ \ \ \ \left(*\right)\end{aligned}
Let $ n\in\mathbb{N}^{*} $, we have the following :
\begin{aligned}\small\left\vert\sum_{k=1}^{n}{\left(-1\right)^{k}\frac{\left(2\pi\right)^{2k}}{\left(2k\right)!}\int_{0}^{1}{x^{2k}\,\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}}-\int_{0}^{1}{\sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\left(2\pi x\right)^{2n}}{\left(2n\right)!}\mathrm{e}^{-2\pi\mathrm{i}x}}\,\mathrm{d}x}\right\vert &\small =\left\vert\int_{0}^{1}{\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k}\frac{\left(2\pi x\right)^{2k}}{\left(2k\right)!}\mathrm{e}^{-2\pi\mathrm{i}x}}\,\mathrm{d}x}\right\vert\\ &\small\leq\int_{0}^{1}{\left\vert\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k}\frac{\left(2\pi x\right)^{2k}}{\left(2k\right)!}\mathrm{e}^{-2\pi\mathrm{i}x}}\right\vert\mathrm{d}x}\\ &\small\leq\int_{0}^{1}{\frac{\left(2\pi x\right)^{2n+2}}{\left(2n+2\right)!}\,\mathrm{d}x}=\frac{\left(2\pi\right)^{2n+2}}{\left(2n+3\right)!}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}
Thus : $$\small \sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\left(2\pi\right)^{2n}}{\left(2n\right)!}\int_{0}^{1}{x^{2n}\,\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}}=\lim_{n\to +\infty}{\sum_{k=1}^{n}{\left(-1\right)^{k}\frac{\left(2\pi\right)^{2k}}{\left(2k\right)!}\int_{0}^{1}{x^{2k}\,\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}}} =\int_{0}^{1}{\sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\left(2\pi x\right)^{2n}}{\left(2n\right)!}\mathrm{e}^{-2\pi\mathrm{i}x}}\,\mathrm{d}x} $$
$$ \text{(In other words we can interchange the sum with the integral)} $$
We get that : \begin{aligned}\sum_{n=1}^{+\infty}{\sum_{k=1}^{2n}{\frac{\left(-1\right)^{n+1}\left(2\pi\right)^{2n}}{\left(2n-k+1\right)!\left(2\pi\,\mathrm{i}\right)^{k}}}}&=\int_{0}^{1}{\sum_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{\left(2\pi x\right)^{2n}}{\left(2n\right)!}\mathrm{e}^{-2\pi\mathrm{i}x}}\,\mathrm{d}x} \\&=\int_{0}^{1}{\left(\cos{\left(2\pi x\right)}-1\right)\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}\\ &\small =\frac{1}{2}\end{aligned}
To get $ \left(*\right) $, we can integrate by parts and prove that : $$ \int_{0}^{1}{x^{2n}\,\mathrm{e}^{-2\pi\mathrm{i}x}\,\mathrm{d}x}=-\frac{\left(2n\right)!}{\left(2\pi\right)^{2n+1}}\sum_{k=1}^{2n}{\frac{\left(2\pi\mathrm{i}\right)^{k}}{k!}} $$
A: Metamorphy has already given the answer of 1/2, but my derivation is different. I'll present a proof if his is not sufficient.  I prove
$$ \sum_{n=0}^\infty \Big( \frac{\Gamma(2n+2,t)}{\Gamma(2n+2)} - 1 \Big) =
\frac{1}{4}(1-e^{-2t} - 2t) $$
Division by $2 \pi i$ and substituting $t=-2\pi i$ gives the answer.
