Prove that $\sum_{n=1}^{\infty }\frac{B_{2n}}{(2n-1)!}=\frac{1}{2}-\frac{1}{(e-1)^2}$

Question

Prove that $$\sum_{n=1}^{\infty }\frac{B_{2n}}{(2n-1)!}=\frac{1}{2}-\frac{1}{(e-1)^2}$$ My idea is to find the Taylor series of $\frac{1}{(e^x-1)^2}$, but it seems not useful.

Any helps, thanks

Answer 1

Exponential generating function of Bernoulli numbers is given by $$ \sum_{n=0}^\infty B_n{x^n\over n!}={x\over e^x-1}\\ $$ which upon differentiation gives $$ \sum_{n=1}^{\infty}B_n{x^{n-1}\over(n-1)!}={e^x-1-xe^x\over (e^x-1)^2} $$ Plugging in $x=\pm 1$ we get $$ \sum_{n=1}^\infty {B_n\over(n-1)!}=-{1\over(e-1)^2},\quad\sum_{n=1}^\infty {(-1)^nB_n\over(n-1)!}={(e-2)e\over(e-1)^2} $$ Adding these up we get $$ 2\sum_{n=1}^{\infty}{B_{2n}\over{(2n-1)!}}=1-{2\over(e-1)^2} $$

Answer 2

An alternative approach is to use the integral representation

$$ B_{2n} = (-1)^{n}4n \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2 \pi x}-1} \, dx.$$

Specifically,

$$ \begin{align}\sum_{n=1}^{\infty} \frac{B_{2n}}{(2n-1)!} &= \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} (-1)^{n-1} 4n \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2 \pi x}-1} \, dx \\ &= 4 \int_{0}^{\infty} \frac{1}{e^{2 \pi x}-1} \sum_{n=1}^{\infty} \frac{n (-1)^{n-1} x^{2n-1}}{(2n-1)!} \, dx. \end{align}$$

But notice that $$ \begin{align} \sin (x) + x \cos(x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n-1}}{(2n-1)!} + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{2n-1}}{(2n-2)!} \\ &= \sum_{n=1}^{\infty} \frac{[1+(2n-1)](-1)^{n-1} x^{2n-1}}{(2n-1)!} \\ &= \sum_{n=1}^{\infty} \frac{2n(-1)^{n-1} x^{2n-1}}{(2n-1)!}. \end{align}$$

So using the fact that $$\int_{0}^{\infty} \frac{\sin ax}{e^{2 \pi x}-1} \, dx = \frac{1}{4} \, \coth \left(\frac{a}{2} \right) - \frac{1}{2a}, $$ we get

$$ \begin{align} \sum_{n=1}^{\infty} \frac{B_{2n}}{(2n-1)!} &= 2 \left[ \int_{0}^{\infty} \frac{\sin x}{e^{2 \pi x}-1} \,dx + \int_{0}^{\infty} \frac{x \cos x}{e^{2 \pi x}-1} \, dx \right] \\ &= 2 \left[\frac{1}{4} \, \coth \left(\frac{1}{2} \right) - \frac{1}{2} + \frac{\mathrm{d}}{\mathrm{d}a} \left(\frac{1}{4} \, \coth \left(\frac{a}{2} \right) - \frac{1}{2a}\right)\Bigg|_{a=1} \right] \\ &= 2 \left[\frac{1}{4} \, \coth \left(\frac{1}{2} \right) - \frac{1}{2} + \frac{1}{2} - \frac{1}{8} \, \text{csch}^{2} \left(\frac{1}{2} \right) \right] \\ &= \frac{1}{2} \, \coth \left(\frac{1}{2} \right) - \frac{1}{4} \, \text{csch}^{2} \left(\frac{1}{2} \right) \\&= \frac{1}{2} \frac{e+1}{e-1} - \frac{1}{4} \frac{4e}{(e-1)^{2}} \\ &= \frac{(e^{2}-1) -2e}{2(e-1)^{2}} \\ &= \frac{(e-1)^{2}-2}{2(e-1)^{2}} \\ &= \frac{1}{2} - \frac{1}{(e-1)^{2}}.\end{align}$$

Answer 3

We can also use the identity $$x\cot\left(x\right)-1=\sum_{n\geq1}\frac{B_{2n}\left(-4\right)^{n}x^{2n}}{\left(2n\right)!},\,\left|x\right|<\frac{\pi}{2} $$ so taking the derivative and manipulating a bit $$x\cot\left(x\right)-\frac{x^{2}}{\sin^{2}\left(x\right)}=\sum_{n\geq1}\frac{B_{2n}\left(-4\right)^{n}x^{2n}}{\left(2n-1\right)!} $$ hence taking $x=\frac{i}{2} $ we get $$\sum_{n\geq1}\frac{B_{2n}}{\left(2n-1\right)!}=\frac{i\cot\left(i/2\right)}{2}+\frac{1}{4\sin^{2}\left(i/2\right)}=\frac{1}{2}-\frac{1}{\left(e-1\right)^{2}}.$$