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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?

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  • $\begingroup$ I have corrected because of the problem with the pole at $x=1$. $\endgroup$ – user90369 Nov 27 '16 at 12:06
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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

$$\sum\limits_{k=1}^\infty z^k\left(1+\frac{1}{k}\right)^k=z\frac{d}{dz}\int\limits_0^\infty \left(\frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})}-1\right) dx =$$ $$=\left(\frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})}-\frac{e^{-W(-z/e)}}{1+W(-z/e)}\right)\frac{x}{1-x}|_0^\infty $$ $$-\int\limits_0^\infty \left( \frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})} - \frac{e^{-W(-z/e)}}{1+W(-z/e)}\right) \frac{1}{(1-x)^2}dx $$ $$=\frac{e^{-W(-z/e)}}{1+W(-z/e)}-1+\int\limits_0^\infty \left(\frac{e^{-W(-z/e)}}{1+W(-z/e)}-\frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})}\right)\frac{dx}{(1-x)^2} $$ $$=\frac{e^{-W(-z/e)}}{1+W(-z/e)}-1+$$ $$+\int\limits_0^1 \left(2\frac{e^{-W(-z/e)}}{1+W(-z/e)}- \frac{e^{-W(-zxe^{-x})}}{1+W(-zxe^{-x})} -\frac{e^{-W\left(-\frac{z}{x} e^{-1/x}\right)}}{1+W\left(-\frac{z}{x} e^{-1/x}\right)}\right)\frac{dx}{(1-x)^2} \hspace{3 mm}.$$

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    $\begingroup$ It could be a good exercise to find how you find it and solve it before you haha :-) $\endgroup$ – Hexacoordinate-C Nov 24 '16 at 18:26
  • $\begingroup$ @Hexacoordinate-C : Why not - would be interesting. :-) $\endgroup$ – user90369 Nov 25 '16 at 14:11

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