is true that $\sum_{n=0}^\infty B_n(-1)^n=\frac{\pi^2}{6}$?

Question

Is true that $\sum_{n=0}^\infty B_n(-1)^n=\frac{\pi^2}{6}$? (where $B_n$is the Bernoulli number)

If this is true, how can I prove it?

Answer 1

Since the even Bernoulli numbers grow as $B_{2n} \approx 4\sqrt{\pi n}(n/(e\pi))^{2n} $, the sum diverges pretty violently.

I will try something else to try to assign a value.

Since $\frac{x}{e^x-1} =\sum_{n=0}^{\infty} \frac{B_n x^n}{n!} $,

we can try a standard "trick" to get rid of the $n!$ in the denominator: multiply by $e^{-xt}$ and integrate from $0$ to $\infty$.

The RHS becomes

$\begin{array}\\ \int_0^{\infty} \sum_{n=0}^{\infty} e^{-xt}\frac{B_n x^n}{n!} dx &= \sum_{n=0}^{\infty} \int_0^{\infty}e^{-xt}\frac{B_n x^n}{n!} dx\\ &= \sum_{n=0}^{\infty}\frac{B_n }{n!} \int_0^{\infty}e^{-xt}x^n dx\\ &= \sum_{n=0}^{\infty}\frac{B_n }{t^nn!} \int_0^{\infty}e^{-xt}(xt)^n dx\\ &= \sum_{n=0}^{\infty}\frac{B_n }{t^{n+1}n!} \int_0^{\infty}e^{-x}(x)^n dx\\ &= \sum_{n=0}^{\infty} B_n t^{-n-1}\\ \end{array} $

The LHS becomes $\int_0^{\infty}\frac{xe^{-xt}}{e^x-1}dx $. If we put $t=-1$, which will make the RHS what we want, this becomes $\int_0^{\infty}\frac{xe^{x}}{e^x-1}dx $. Unfortunately, this integral diverges.

However, putting the indefinite integral into Wolfy, it says that the indefinite integral is $\int (x e^x)/(e^x-1) dx = Li_2(e^x)+x \log(1-e^x) $ (where $Li_2$ is the dilogarithm) and that the expansion about $x=0$ is $\text{(an imaginary expression)}+(\pi^2/6+x+O(x^2)) $ and the expansion about $x = \infty$ is $x \log(1-e^x)-\pi^2/6 $.

These give some justification for thinking that your value might has some validity for some method of summing divergent series.