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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"

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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.

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  • $\begingroup$ So in the second formula I choose x to be -1 in this case. $\endgroup$ – Antal Korpa May 23 '15 at 6:49
  • $\begingroup$ "Li" means the logarithmic integral funtion? How can I calculate the value of the integral? $\endgroup$ – Antal Korpa May 23 '15 at 6:50
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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 $$A(2,1) = \zeta(3) - \dfrac{\zeta(2)\log(2)}2$$

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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.

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  • $\begingroup$ A derrivation would be nice along that. $\endgroup$ – cirpis May 22 '15 at 18:11
  • $\begingroup$ Some of these symbolic algorithms have "certification methods" attached to them -- e.g., the Wilf-Zeilberger method. There are some add-on packages for Mathematica that show such certificates. This is probably not what you mean exactly by "derivation method". But the fact is that it does become a fully automated proof. Some computer algorithms allow such certifications, others do not. Once should always check numerically that the symbolic result returned agrees with the numeric approximation, of course. $\endgroup$ – stan wagon May 22 '15 at 19:12

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