Proving that $~\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{k^2}~H_k~H_{k-1}=\frac{3}{16}~\zeta(4)$ To show that 
$$\sum\limits_{k=2}^{\infty} \frac{(-1)^{k}}{k^{2}} \, \left(1+\frac{1}{2}+...+\frac{1}{k}\right) \cdot \left(1+\frac{1}{2}+...+\frac{1}{k-1}\right) = \frac{3}{16}\zeta(4).$$
I came across this when trying to solve a problem from the current edition of the American Mathematical Monthly. Is there some easy way to show this? I checked numerically that this series does converge to the value of $\frac{3}{16}\zeta(4)$.
Note: An alternate form, with $H_{n}$ being the harmonic numbers, is:
$$ \sum\limits_{k=2}^{\infty} \frac{(-1)^{k}}{k^{2}} \, H_{k} \, H_{k-1} = \frac{3}{16}\zeta(4). $$
 A: Although I have seen too few proofs in this field to be able to compare, this approach might be interesting.
We transform the sum to a fourfold integral which Mathematica can solve immediately. I hope it should be possible to solve the integral "mathematically" as well, which would then complete the proof.
We have to calculate
$$s=\sum _{k=2}^{\infty } \frac{(-1)^k}{k^2} H(k) H(k-1) $$
Writing
$$\frac{1}{n^2}=\int_0^1 \frac{1}{x}\,dx \int_0^x y^{n-1} \, dy $$
$$\frac{1}{n}=\int_0^1 r^{n-1} \, dr$$
and
$$H(k)=\sum _{n=1}^k \frac{1}{n}=\int_0^1 \left(\sum _{n=1}^k r^{n-1}\right) \, dr=\int_0^1 \frac{1-r^k}{1-r} \, dr$$
the sum $s$ below the integrals becomes
$$si=\frac{1}{x(1-r)(1-s)}\sum _{k=2}^{\infty } (-1)^k \left(1-r^k\right) \left(1-s^{k-1}\right) y^{k-1}$$
Which evaluates to
$$si = \frac{y \left(r^2 s^2 y+r^2 s-r^2 y-r^2-r s^2 y+r y-s+1\right)}{(1-r) (1-s) x (y+1) (r y+1) (s y+1) (r s y+1)}$$
Now the integral to be evaluated is
$$s4 = \int _0^1 dx\int _0^x dy\int _0^1 dr\int _0^1 ds \; si$$
Mathematica finds immediately
$$s4 = \frac{\pi ^4}{480} $$
Since 
$$\zeta (4)=\frac{\pi ^4}{90}$$
and 
$$\frac{90}{480} = \frac{3}{16} $$
we have finally
$$s = \frac{3}{16} \zeta(4)$$
A: Bonus
$$S=\sum_{k=2}^\infty\frac{(-1)^k}{k^2}H_kH_{k-1}=\sum_{k=1}^\infty\frac{(-1)^k}{k^2}H_kH_{k-1}=\sum_{k=1}^\infty\frac{(-1)^kH_k^2}{k^2}-\sum_{k=1}^\infty\frac{(-1)^kH_k}{k^3}$$
Since $S=\frac3{16}\zeta(4)$ and $\sum_{k=1}^\infty\frac{(-1)^kH_k}{k^3}$ $=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$
then 
$$\sum_{n=1}^{\infty}\frac{(-1)^kH_k^2}{k^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$
