real analysis - Is $sumlimits_{n = 1}^infty (-1)^nfrac{H_n}{n}=frac{6ln^2(2)-pi^2}{12}$ a valid identity?

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.

QUESTIONS:

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