A series involving harmonic numbers Does anyone know the exact value of this:
$$
\sum_{k=1}^{\infty} (-1)^k\frac{H_k}{k}
$$
or this:
$$
\sum_{k=1}^{\infty} (-1)^k\frac{H_k^{(2)}}{k}
$$
Thanks!

Thanks again for the answers! I found very interesting that the integral gives exact values up to r=3 but from 4 this integral gives not exact values:
$$
\int_{-1}^0 \frac{Li_4(t)}{t(1-t)} \mathrm{d}t
$$ 
because wolfram says that "no results found in terms of standard mathematical functions"
 A: From my answer in my previous avatar, denoting
$$A(p,q) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}H_k^{(p)}}{k^q}$$
we have
$$A(1,1) = \dfrac{\zeta(2)-\log^2(2)}2$$
and
$$A(2,1) = \zeta(3) - \dfrac{\zeta(2)\log(2)}2$$
A: Hint:
One of the usual tactics is to exploit the generating function for harmonic numbers:
$$\sum_{n=1}^{\infty}H_{n}^{(r)}t^n=\frac{\operatorname{Li}_{r}{\left(t\right)}}{1-t};~~~\small{\left[\left|t\right|<1\right]}.$$
Dividing both sides by $t$ and integrating from $0$ to $x$, we have:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_{n}^{(r)}x^{n}}{n}
&=\int_{0}^{x}\frac{\operatorname{Li}_{r}{\left(t\right)}}{t\left(1-t\right)}\,\mathrm{d}x.\\
\end{align}$$
Thus, to find the values of your sums, you simply set $x$ to an appropriate value and solve the integral on the RHS.
A: The answer to the first is (6*Log[2]^2 - Pi^2)/12, obtained by calling on algorithms for symbolic summation included in Mathematica. 
Input:  Sum[((-1)^k*HarmonicNumber[k])/k, {k, 1, Infinity}]
Output: (-Pi^2 + 6*Log[2]^2)/12
I am not sure what your superscript (2) means.
