Value of the series $\sum_\limits{n=1}^{\infty}\left(\frac{n+1}{n\cdot 2\pi}\right)^ n$ What is the exact value of the series
$$S=\sum_{n=1}^{\infty}\left(\frac{n+1}{n\cdot 2\pi}\right)^ n=\sum_{n=1}^{\infty}\left(\frac{1}{ 2\pi}\left(1+\frac{1}{n}\right)\right)^n?$$
By the obvious inequalities $$1\leq\left(1+\frac{1}{n}\right)\leq 2$$
for every $n\in\mathbb{N}$ one gets $$0.18927...\approx\frac{1}{1-\frac{1}{2\pi}}-1\leq\sum_{n=1}^{\infty}\left(\frac{1}{ 2\pi}\right)^n\leq S\leq\sum_{n=1}^{\infty}\left(\frac{1}{ \pi}\right)^n=\frac{1}{1-\frac{1}{\pi}}-1\approx 0.46694...$$
and numerically $$S\approx 0.38672...,$$
but I'm stuck at determining a closed expression for this series. 
Any ideas or solutions for this?
 A: An integral representation is 
$$\sum\limits_{k=1}^\infty z^k\left(1+\frac{1}{k}\right)^k=\frac{e^{-W(-z/e)}}{1+W(-z/e)}-1+$$ $$+\int\limits_0^1 \left(2\frac{e^{-W(-z/e)}}{1+W(-z/e)}- \frac{e^{- W\left(-zxe^{-x}\right) }}{1+W\left(-zxe^{-x}\right) } -\frac{e^{- W\left(-\frac{z}{x}e^{-\frac{1}{x}}\right) }}{1+ W\left(-\frac{z}{x}e^{-\frac{1}{x}}\right)  }\right)\frac{1}{(1-x)^2}dx$$ 
for $|z|<1$ and with $W(x)$ as the main branch of the Lambert-W-function. 
For your series set $\displaystyle z:=\frac{1}{2\pi}$ .

For the calculations we need the following basic informations : 
$(1)\hspace{3 mm} \displaystyle\int\limits_0^\infty (xe^{-x})^k dx=\frac{(k-1)!}{k^k}$
$(2)\hspace{3 mm} \displaystyle\frac{e^{-W(-z)}}{1+W(-z)} =\sum\limits_{k=0}^\infty \frac{(z(k+1))^k}{k!}$ uniformly convergent for $\displaystyle |z|<\frac{1}{e}$
Calculations:
Part $(A)$:
Be $z\to zxe^{-x}$ with $|z|<1$ and $x\in\mathbb{R}_0^+$ .
$$\frac{e^{-W(-zxe^{-x}) }}{1+W(-zxe^{-x})}-1=\sum\limits_{k=1}^\infty (xe^{-x})^k\frac{k^k}{k!}z^k\left(1+\frac{1}{k}\right)^k$$ 
and therefore 
$$\int\limits_0^\infty (\frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})}-1) dx =\int\limits_0^\infty (\sum\limits_{k=1}^\infty (xe^{-x})^k\frac{k^k}{k!}z^k(1+\frac{1}{k})^k) dx $$ $$= \sum\limits_{k=1}^\infty (\int\limits_0^\infty (xe^{-x})^k dx)\frac{k^k}{k!}z^k(1+\frac{1}{k})^k =\sum\limits_{k=1}^\infty \frac{z^k}{k}(1+\frac{1}{k})^k $$
Part $(B)$ with differentiable functions $f$ and $g$ where $g(1)\neq 0$ and $g'(1)=0$, and with the interchangeability of derivative and integration:
It’s $\enspace\displaystyle \frac{d}{dz}f(zg(x))=f'(zg(x))g(x)\enspace$ and $\enspace\displaystyle \frac{d}{dx}f(zg(x))=f'(zg(x))zg’(x)\enspace$ and therefore 
$$z\frac{d}{dz}\int\limits_a^b f(zg(x))dx= \int\limits_a^b f’(zg(x))zg’(x)\frac{g(x)}{g’(x)}dx=$$
$$=(f(zg(x))-f(zg(1)))\frac{g(x)}{g’(x)}|_{x=a}^{x=b}- \int\limits_a^b (f(zg(x))-f(zg(1)))\left(\frac{g(x)}{g’(x)}\right)’dx $$  
Part $(C)$:
With $\enspace\displaystyle f(z):= \frac{e^{-W(-z)}}{1+W(-z)}-1\enspace$ and $\enspace\displaystyle g(x):=xe^{-x}\enspace$ and therefore $\enspace\displaystyle \left(\frac{g(x)}{g’(x)}\right)’=\frac{1}{(1-x)^2}\enspace$ it follows 
$$\sum\limits_{k=1}^\infty z^k\left(1+\frac{1}{k}\right)^k=z\frac{d}{dz}\int\limits_0^\infty \left(\frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})}-1\right) dx =$$
$$=\left(\frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})}-\frac{e^{-W(-z/e)}}{1+W(-z/e)}\right)\frac{x}{1-x}|_0^\infty $$ $$-\int\limits_0^\infty \left( \frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})} - \frac{e^{-W(-z/e)}}{1+W(-z/e)}\right) \frac{1}{(1-x)^2}dx $$
$$=\frac{e^{-W(-z/e)}}{1+W(-z/e)}-1+\int\limits_0^\infty \left(\frac{e^{-W(-z/e)}}{1+W(-z/e)}-\frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})}\right)\frac{dx}{(1-x)^2} $$
$$=\frac{e^{-W(-z/e)}}{1+W(-z/e)}-1+$$ $$+\int\limits_0^1 \left(2\frac{e^{-W(-z/e)}}{1+W(-z/e)}- \frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})} -\frac{e^{-W\left(-\frac{z}{x} e^{-1/x}\right)}}{1+W\left(-\frac{z}{x} e^{-1/x}\right)}\right)\frac{dx}{(1-x)^2} \hspace{3 mm}.$$
