# 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.

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 $$\frac{2}{1+1}+\frac{2}{1+3}+\ldots+\frac{2}{2h} = H_h.$$