How can we one show that $\sum_{i=1}^{\infty}\sum_{j=1}^{2k}(-1)^{j-1}{2\over i+j}=H_k?$ Given the double sums $(1)$

$$\sum_{i=1}^{\infty}\sum_{j=1}^{2k}(-1)^{j-1}{2\over i+j}=\color{blue}{H_k}\tag1$$
  Where $H_k$ is the n-th harmonic number

How can one prove $(1)$?
Rewrite $(1)$ as
$$\sum_{i=1}^{\infty}\left({2\over i+1}-{2\over i+2}+{2\over i+3}-\cdots+{2\over i+k}\right)\tag2$$
Rewrite $(2)$ as
$$\sum_{i=1}^{\infty}\left({2\over (i+1)(i+2)}+{2\over (i+3)(i+4)}+{2\over (i+5)(i+6)}+\cdots+{2\over (i+k)(i+k+1)}\right)\tag3$$
Help required, not sure what is the next  step. Thank you.
 A: There is some issue: if $k$ is odd, the main term of $(2)$ behaves like $\frac{C}{i}$, leading to a divergent sum.
On the other hand, if $k=2h$ is even then $(2)$ is an absolutely convergent  telescopic sum:
$$ \sum_{i\geq 1}\left(\frac{2}{i+1}-\frac{2}{i+2}+\frac{2}{i+3}-\frac{2}{i+4}+\ldots+\frac{2}{i+2h-1}-\frac{2}{i+2h}\right) $$
that clearly equals
$$ \frac{2}{1+1}+\frac{2}{1+3}+\ldots+\frac{2}{2h} = H_h. $$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{i = 1}^{\infty}\sum_{j = 1}^{2k}\pars{-1}^{\,j - 1}{2 \over i + j} & =
2\sum_{i = 1}^{\infty}
\sum_{j = 1}^{2k}\pars{-1}^{\,j - 1}\int_{0}^{1}x^{i + j - 1}\,\dd x =
2\sum_{i = 1}^{\infty}x^{i}
\int_{0}^{1}\sum_{j = 1}^{2k}\pars{-x}^{\,j - 1}\,\dd x
\\[5mm] & =
2\sum_{i = 1}^{\infty}x^{i}
\int_{0}^{1}{\pars{-x}^{2k} - 1 \over -x - 1}\,\dd x =
2\int_{0}^{1}{1 - x^{2k} \over 1 + x}\sum_{i = 1}^{\infty}x^{i}\,\dd x
\\[5mm] & =
2\int_{0}^{1}{x - x^{2k + 1} \over 1 - x^{2}}\,\dd x
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
\int_{0}^{1}{x^{k} - 1 \over x - 1}\,\dd x = \int_{0}^{1}
\sum_{n = 1}^{k}x^{n - 1}\,\dd x
\\[5mm] & = \sum_{n = 1}^{k}{1 \over n} = \bbx{\ds{H_{k}}}
\end{align}
