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.
 A: 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, $\frac11+\dots+\frac1p.$
A: Using Mathmatica, we have
$$S=\sum_{k = 1}^\infty  \frac{(-1)^{k - 1}}{k} \sum_{n = 0}^\infty \frac{1}{k \, 2^n + 5}$$
$$a_k=\sum_{n = 0}^\infty \frac{1}{k \, 2^n + 5}=\frac 1{10}+\frac{\psi _2\left(-\frac{\log \left(-\frac{5}{k}\right)}{\log
   (2)}\right)+\log \left(-\frac{5}{k}\right)}{5\log (2)}$$ the where appears the q-digamma function.
Computed to large accuracy
$$S=0.456666666666666\sim\frac{137}{300}$$ as already reported by  @Robert Israel.
For large values of $k$, an expansion gives
$$a_k=\sum_{p=1}^\infty \frac {b_p}{c_p} k ^{-p}$$ where the numerators $b_p$ make the sequence
$$\{7,-425,117025,-139810125,692736660625,-13962052416203125\}$$ (which is not recognized by $OEIS$) and the denominators $c_p$
$$\{4,64,4096,1048576,1073741824,4398046511104\}$$ that is to say $c_p=2^{p(p+1)}$.
A: The sum is $$\sum^\infty_{k=1}\frac1{k\,(k+5)}=\frac15\,\sum^\infty_{k=1}\left(\frac1k-\frac1{k+5}\right)=\frac15\,\left(1+\frac12+\frac13+\frac14+\frac15\right)=\frac{137}{300}.$$
Cf. Evaluate $S=\sum_{k=1}^\infty \frac{(-1)^{k-1}} k \sum_{n=0 }^\infty \frac 1 { k \cdot 2^n+1 } $
