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