A while ago I asked a question about an identity that I found while playing with series involving harmonic numbers. However, since the methods I used were not the focus of my question, I never explained how I got this result. So here it goes...

NOTE: It may be of help to work backwards through this problem instead. This is just how I did it.

We start by defining $H_n$ as the $n^{th}$ harmonic number. We also take note of the identities:

$$ \sum_{n=1}^\infty \frac{(-1)^n}{n} = -\ln(2)$$ $$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$

Now we note that:

$$ \tag{1}H_n = \sum_{k=1}^\infty \frac{1}{k} - \frac{1}{k+n} = \sum_{k=1}^\infty \frac{n}{k(k+n)}$$

Since $\sum_{k=1}^\infty \frac{1}{k} - \frac{1}{k+n}$ is a telescoping series.

Dividing by $n$ we get: $$\tag{2} \frac{H_n}{n} = \sum_{k=1}^\infty \frac{1}{k(k+n)}$$

Since $ \sum_{n=1}^\infty\frac{H_n}{n}$ is obviously divergent, I took a look at the alternating series:

$$ \tag{3}\sum_{n=1}^\infty(-1)^n\frac{H_n}{n} = \sum_{n=1}^\infty(-1)^n\left(\sum_{k=1}^\infty \frac{1}{k(k+n)}\right)$$

We must now multiply this series by 2 and add $\sum_{n=1}^\infty\frac{1}{n^2} $ which is necessary for the next step.

So we have:

$$\begin{align}\tag{4}\sum_{n=1}^\infty\frac{1}{n^2} + 2 \sum_{n=1}^\infty(-1)^n\left(\sum_{k=1}^\infty \frac{1}{k(k+n)}\right) & = \left(\frac{1}{1}\cdot\frac{1}{1} + \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{3}\cdot\frac{1}{3}+\cdots\right) \\ & -2 \left(\frac{1}{1}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{4}+\cdots\right) \\ & +2 \left(\frac{1}{1}\cdot\frac{1}{3} + \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{3}\cdot\frac{1}{5}+\cdots\right) \\ & -2 \left(\frac{1}{1}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{5} + \frac{1}{3}\cdot\frac{1}{6}+\cdots\right) \\ & + \cdots \end{align}$$

With a very careful (and possibly illegal) rearrangement of this expanded sum, we can finally obtain: $\\$ (NOTE: It may be easier to work backwards through this step)

$$ \begin{align} \tag{5}\sum_{n=1}^\infty\frac{1}{n^2} + 2 \sum_{n=1}^\infty(-1)^n\left(\sum_{k=1}^\infty \frac{1}{k(k+n)}\right) &= \left(-\frac{1}{1} + \frac{1}{2} - \frac{1}{3} + \cdots\right)\left(-\frac{1}{1} + \frac{1}{2} - \frac{1}{3} + \cdots\right) \\ & = \ln^2(2)\end{align}$$

Thus, we have:

$$\tag{6}\sum_{n=1}^\infty(-1)^n\frac{H_n}{n} = \frac{6\ln^2(2) - \pi^2}{12} = -0.5822...$$

Here are 3 graphs from computing the actual value of the sum. The first graph is of the difference between the actual sum and $\frac{6*ln^2(2)-\pi^2}{12}$ for k up to 14000. As you can see it seems to converge on 0. You can ignore the 2nd and 3rd graphs.


Is the method used from step 4-5 (or vise versa) legal even though it involves conditionally convergent series? If not, is there any restrictions and/or rearrangements I can use to make it legal?

Is there any known generalizations for $\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^q}$ that extend this result?

Recommended readings that cover this topic would be greatly appreciated as well.

Thanks, Dom

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    $\begingroup$ $H_n$ depends on n, so it cannot be a limit $\endgroup$ – Alex Nov 8 '16 at 9:24
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    $\begingroup$ I would use a generating functions approach: It is well known that the harmonic numbers are generated by $$ g(x)=-\frac{\log(1-x)}{1-x}=\sum_{n\geq 1} H_nx^n $$ now denote the sum in question by $S$. It is easy to see that it can be obtained from the function $$ S(x)=\int^x \frac{g(x')}{x'}dx=\sum_{n\geq 1} \frac{H_n}{n}x^n $$ via $S=S(-1)$ To calulate the integral we perform a partial fraction decompositon $$ S(x)=-\int^x dx'\frac{\log(1-x')}{x'}-\int^x dx'\frac{\log(1-x')}{1-x'} $$ ... $\endgroup$ – tired Nov 8 '16 at 9:54
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    $\begingroup$ ...the second integral is trivially integrated in terms of elementary functions yielding $$ S(x)=-\int^x dx'\frac{\log(1-x')}{x'}+\frac{1}{2}\log(1-x)^2 $$ for the remaining integral we integrate the taylor series of $\log(1-x)/x$ term by term $$ S(x)=\sum_{m\geq1}\frac{x^m}{m^2}+\frac{\log(1-x)^2}{2} $$ and therfore $$ S=S(-1)=\sum_{m\geq1}\frac{(-1)^m}{m^2}+\frac{\log^2(2)}{2}=\\ $$ or >$$ S=\frac{\log(2)^2}{2}-\frac{\pi^2}{12} $$ as announced $\endgroup$ – tired Nov 8 '16 at 9:54
  • $\begingroup$ Thank you very much! If you make this response an answer I will certainly confirm it. $\endgroup$ – Dom Nov 8 '16 at 21:33
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    $\begingroup$ i think my comment is fine...i just want to mention that this also works for sums with higher powers of $n$ in the denominator. for example $$\sum_{n\geq 1}\frac{H_n x^n}{n^2}=\int^x \frac{dx'}{x'} \int^{x'} \frac{dx''g(x'')}{x''}$$. $\endgroup$ – tired Nov 9 '16 at 7:45