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"
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 $$A(2,1) = \zeta(3) - \dfrac{\zeta(2)\log(2)}2$$
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.
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.
