<s>[INST] Evaluating double sum $\sum_{k = 1}^\infty \left( \frac{(-1)^{k - 1}}{k} \sum_{n = 0}^\infty \frac{1}{k \cdot 2^n + 5}\right)$ 
Find $$\sum_{k = 1}^\infty \left( \frac{(-1)^{k - 1}}{k} \sum_{n = 0}^\infty \frac{1}{k \cdot 2^n + 5}\right)$$

So far, I've gotten that the sum of the left is equal to $\log(2),$ meaning we have to evaluate $\displaystyle \sum_{n=0}^{\infty} \frac{\log(2)}{k\cdot2^n+5},$ but I don't know how to proceed. I don't think it's geometric or we can use Partial Fraction Decomposition on it.
 [/INST] The inner sum is bounded by $\frac{2}{k},$ so the full sum is absolutely convergent, which means we can rearrange and combine terms.
If $m=2^i p$ for $p$ odd, the coefficient of $\frac1{m+5}$ is $$\frac{1}{p}-\left(\frac{1}{2p}+\frac{1}{4p}+\cdots +\frac{1}{2^ip} \right)=\frac{1}{m}.$$
So your sum is: $$\sum_{m=1}^{\infty}\frac{1}{m(m+5)}\tag1$$
Since $$\frac1{m(m+5)}=\frac15\left(\frac1m-\frac1{m+5}\right)$$
$(1)$ is equal to:
$$\frac{1}5\left(\frac{1}1+ \frac{1}2+ \frac{1}3+ \frac{1}4+ \frac{1}5\right)$$

This gives us more generally, for any positive integer $p:$ $$\sum_{k = 1}^\infty \left( \frac{(-1)^{k - 1}}{k} \sum_{n = 0}^\infty \frac{1}{k \cdot 2^n + p}\right)=\frac{1}{p}H_p,$$ where $H_p$ is the harmonic number, $\frac11+\dots+\frac1p.$
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