$\sum{\frac{n}{{{e^{2\pi n}} - 1}}}$型的级数求解 - Eufisky - The lost book
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$\sum{\frac{n}{{{e^{2\pi n}} - 1}}}$型的级数求解

Eufisky posted @ 2014年7月24日 20:24 in 数学分析 with tags 级数 Mellin变换 , 5439 阅读
两个结论:\[\left\{ \begin{array}{l}\sum\limits_{n = 1}^\infty  {\frac{n}{{{e^{2\pi n}} - 1}}}  = \frac{1}{{24}} - \frac{1}{{8\pi }}\\\sum\limits_{n = 0}^\infty  {\frac{{2n + 1}}{{{e^{\pi \left( {2n + 1} \right)}} + 1}}}  = \frac{1}{{24}}\end{array} \right..\]

Proof.What you require here are the Eisenstein series. In particular the evaluation of

 
$$E_2(\tau) = 1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}},$$
 
at $\tau = i. $ Rearrange to get
 
$$\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau} } = \frac{1}{24}(1 – E_2(i) ).$$
 
See Lambert series for additional information.
 
The function
 
$$G_ 2(\tau) = \zeta(2) \left(1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}} \right)=\zeta(2)E_2(\tau)$$
 
satisfies the quasimodular transformation
 
$$G_ 2\left( \frac{a\tau+b}{c\tau+d} \right) =(c\tau+d)^2G_ 2(\tau) - \pi i c (c\tau + d).$$
 
And so with $a=d=0,$ $c=1$ and $b=-1$ we find $G_ 2(i) = \pi/2.$ Therefore
 
$$E_2(i) = \frac{ G_ 2( i)}{ \zeta(2)} = \frac{\pi}{2}\frac{6}{\pi^2} = \frac{3}{\pi}.$$
 
Hence we obtain
 
$$\sum_{n=1}^\infty \frac{n}{e^{2\pi n} – 1} = \frac{1}{24} - \frac{1}{8\pi},$$
 
as given in the comment to the question by Slowsolver.
 
 
There is a very nice generalisation of the sum in the question.
 
For odd $ m > 1 $ we have
 
$$\sum_{n=1}^\infty \frac{n^{2m-1} }{ e^{2\pi n} -1 } = \frac{B_{2m}}{4m},$$
 
where $B_k$ are the Bernoulli numbers defined by
 
\[\frac{z}{{{e^z} - 1}} = \sum\limits_{k = 0}^\infty  {\frac{{{B_k}}}{{k!}}} {z^k}\;\;\; for |z|<2\pi .\]
另解:
$$ \displaystyle\mathcal{M}\Big[\frac{x}{e^{2\pi x}-1}] = \int_{0}^{\infty} \frac{x^{s}}{e^{2 \pi x}-1} \ dx = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) $$
 
So $$\displaystyle  \frac{x}{e^{2\pi x}-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) x^{-s}\ ds $$
 
which implies $$ \displaystyle \sum_{n=1}^{\infty}\frac{n}{e^{2\pi n}-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s)\ ds $$
 
The integrand has poles at $s=-1, s=0$, and $s=1$ (and removable singularities at $s= -2, -3, -4, \ldots$)
 
I'm going to close the contour with a rectangle that has vertices at $$-i \infty, \frac{3}{2} - i \infty, \frac{3}{2} + i \infty, and \;i \infty$$ and is indented at the origin
 
$\Gamma(s)$ decays rapidly as $\text{Im} (s) \to \pm \infty$. So the integral goes to zero along the top and bottom of the rectangle.
 
And on the imaginary axis, the integrand is odd.
 
So $$\displaystyle \int_{\frac{3}{2}-i\infty}^{\frac{3}{2}+i\infty}(2\pi)^{-s}\Gamma(s+1)\zeta(s+1)\zeta(s)\ ds-\pi i \text{Res}[f,0] = 2\pi i\text{Res}[f,1]$$
 
where $$\displaystyle f(s) = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s) $$.
 
$$ \displaystyle \text{Res}[f,0] = \lim_{s \to 0} s (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) = \lim_{s\to 0} s\zeta(s+1) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s)$$
 
$$= 1\Big(\frac{1}{2 \pi} \Big)(1)\zeta(0)=-\frac{1}{4 \pi} $$
 
$$ \displaystyle\text{Res}[f,1] = \lim_{s \to 1} (s-1) (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) $$
 
$$= \displaystyle \lim_{s\to 1}(s-1)\zeta(s) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)= 1\Big(\frac{1}{4 \pi^{2}}\Big)(1)\Big(\frac{\pi^{2}}{6}\Big) =\frac{1}{24} $$
 
Therefore,  $$\displaystyle \sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1} = \frac{1}{2 \pi i} \Big( 2 \pi i  (\frac{1}{24}) +  \pi i (\frac{-1}{4 \pi}) \Big) = \frac{1}{24} - \frac{1}{8 \pi}.$$
 
Poof.We will use the Mellin transform technique. Recalling the Mellin transform and its inverse
 
$$ F(s) =\int_0^{\infty} x^{s-1} f(x)dx, \quad\quad f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} F(s)\, ds.  $$
 
 
Now, let's consider the function
 
$$ f(x)= \frac{x}{e^{\pi x}+1}. $$
 
Taking the Mellin transform of $f(x)$, we get
 
$$ F(s)={\pi }^{-s-1}\Gamma  \left( s+1 \right)  \left(1- {2}^{-s} \right) \zeta  \left( s+1 \right),$$
 
where $\zeta(s)$ is the zeta function. Representing the function in terms of the inverse Mellin Transform, we have
 
$$ \frac{x}{e^{\pi x}+1}=\frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma  \left( s+1 \right)  \left( 1-{2}^{-s} \right) \zeta  \left( s+1 \right) x^{-s}ds. $$
 
Substituting $x=2n+1$ and summing yields
 
$$\sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{2\pi i}\int_{C}{\pi}^{-s-1}\Gamma  \left( s+1 \right)\left(1-{2}^{-s} \right) \zeta\left( s+1 \right) \sum_{n=0}^{\infty}(2n+1)^{-s}ds$$
 
$$ = \frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma  \left( s+1 \right)  \left(1-{2}^{-s} \right)^2\zeta\left( s+1 \right) \zeta(s)ds.$$
 
Now, the only contribution of the poles comes from the simple pole $s=1$ of $\zeta(s)$ and the residue equals to $\frac{1}{24}$. So, the sum is given by
 
$$ \sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{24} $$
 
Notes: 1)  
 
 
$$ \sum_{n=0}^{\infty}(2n+1)^{-s}= \left(1- {2}^{-s} \right) \zeta  \left( s \right).  $$
 
2) The residue of the simple pole $s=1$, which is the pole of the zeta function, can be calculated as
 
$$ r = \lim_{s=1}(s-1)({\pi }^{-s-1}\Gamma  \left( s+1 \right)  \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) \zeta(s))$$
 
$$  =  \lim_{s\to 1}(s-1)\zeta(s)\lim_{s\to 1}  {\pi }^{-s-1}\Gamma  \left( s+1 \right)  \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right)  = \frac{1}{24}. $$
 
For calculating the above limit, we used the facts
 
$$ \lim_{s\to 1}(s-1)\zeta(s)=1, \quad \zeta(2)=\frac{\pi^2}{6}. $$
 
3) Here is the technique for computing the Mellin transform of $f(x)$. 
Using the change of variables $u=-\ln(x)$ and the identity
 
$$ \int_{0}^{\infty}\frac{u^{s-1}}{e^u -1}=\zeta{(s)}\Gamma{(s)} $$
 
we reach to the deisred result
 
$$ \int_0^1 \frac{\ln x }{x-1}= \int_{0}^{\infty}\frac{u}{e^u -1}=\zeta{(2)}\Gamma{(2)} =\sum_{n=1}^\infty \frac{1}{n^2}. $$
 
Note that,
 
$$ \int_{0}^{\infty}\frac{u^{s-1}}{e^u - 1}=\int_{0}^{\infty}\frac{u^{s-1}}{e^u}(1-e^{-u})^{-1}= \sum_{n=0}^{\infty} \int_{0}^{\infty}{u^{s-1}e^{-(n+1)u}}$$ 
 
$$= \sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \int_{0}^{\infty}{y^{s-1}e^{-y}}= \sum_{n=1}^{\infty}\frac{1}{n^s} \Gamma(s)= \zeta(s) \Gamma(s).$$
 
另一个积分公式:
 
 
If $Re(s)>1,Re(q)>0$,define\[\zeta(s,q)=\sum_{n=0}^{\infty}\frac1{(q+n)^s}.\],then the function has an integral representation in terms of the Mellin transform as \[\zeta(s,q)=\frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}e^{-qt}}{1-e^{-t}}dt.\]
 
 
 
 
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