A Double sum with something weird between the summations $$\sum_{k=1}^{\infty} \Bigg[ \dfrac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty} \dfrac{300}{k \cdot 2^n +5}\Bigg]= ?$$
I've never seen such a complex summation before. Can anyone help me? It would also be extremely helpful if anyone could tell me if they've seen this type of summation, and where to find similar problems. Has it appeared in any mathematics competitions, or are there similar questions for me to attempt? 
Thanks in advance!
*note: This is Problem 29 from the Chen Jingrun's Cup Secondary School Mathematics Competition 2018. The answer is 137.
 A: This becomes easier by converting to an integral to get that weird linear term out of the denominator. Jack gave the quick version in the comments, but here's the more detailed explanation.
First, introduce an integral to convert from a zeta-like series to a power series:
\begin{multline}
\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty \frac{300}{2^nk+5} = 300\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty\int_0^1 x^{2^nk+4}dx
\\= 300\int_0^1x^4\left[\sum_{n=0}^\infty\sum_{k=1}^\infty (-1)^{k-1}\frac{x^{2^nk}}{k}\right] dx = 300\int_0^1x^4\left[\sum_{n=0}^\infty\ln\left(1+x^{2^n}\right)\right] dx,
\end{multline}
where we used the series expansion $\ln(1+z) = \sum_{k=0}^\infty (-1)^{k-1}z^k/k$ to do the $k$ summation.
Next, evaluate the $n$ series using
$$
\sum_{n=0}^\infty \ln(1+x^{2^n}) = \ln\left[\prod_{n=0}^\infty \left(1+x^{2^n}\right)\right] = \ln\left[\sum_{m=0}^\infty x^m\right] = \ln\left(\frac{1}{1-x}\right) = -\ln(1-x).
$$
The identity with the infinite product can be seen by considering expanding each exponent $m$ as sums of powers of $2$. So now we've got our sum in terms of a nice integral,
$$
\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty \frac{300}{2^nk+5} = -300\int_0^1x^4\ln(1-x)dx
$$
Last step is to use the identity $\int_0^1 x^n\ln(x)dx =-(n+1)^{-2}$ to get 
$$
-300\int_0^1x^4\ln(1-x)dx = -300\int_0^1(1-x)^4\ln(x)dx = 300\left(1 - \frac{4}{4} + \frac{6}{9}-\frac{4}{16}+\frac{1}{25}\right) = 137
$$
So, in conclusion, 
$$
\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty \frac{300}{2^nk+5} =137
$$
