MBMT 2020 – Problem 44

Solution: For the denominator, we have: $$\begin{align} \sum_{n=1}^{\infty}a_n & = \sum_{n=1}^{\infty}\sum_{d|n}\dfrac{1}{2^{d+\frac{n}{d}}} = \sum_{d=1}^{\infty}\sum_{n\ge1,d|n}\dfrac{1}{2^{d+\frac{n}{d}}} \\ & = \sum_{d=1}^{\infty}\dfrac{1}{2^d}\sum_{k=1}^{\infty}\dfrac{1}{2^k} \\ &=\sum_{d=1}^{\infty}\dfrac{1}{2^d}\cdot 1=1 \\ \end{align} $$

For the numerator, we have: $$\begin{align} \sum_{n=1}^{\infty}na_n & = \sum_{n=1}^{\infty}n\sum_{d|n}\dfrac{1}{2^{d+\frac{n}{d}}} = \sum_{d=1}^{\infty}\sum_{n\ge1,d|n}\dfrac{n}{2^{d+\frac{n}{d}}} \\ & = \sum_{d=1}^{\infty}\sum_{n\ge1,d|n}\dfrac{d\cdot\dfrac{n}{d}}{2^{d+\frac{n}{d}}} = \sum_{d=1}^{\infty}\dfrac{d}{2^d}\sum_{k=1}^{\infty}\dfrac{k}{2^k} \\ & =\sum_{d=1}^{\infty}\dfrac{d}{2^d}\cdot 2=2\cdot 2=4 \\ \end{align} $$

Therefore, the answer to the question is: $\boxed{4}$

Note For the sum of the arithmetico–geometric series: $S=\sum_{n=1}^{\infty}\dfrac{n}{2^n}$, we have $$\begin{align} S&=2\cdot S – S =2\sum_{n=1}^{\infty}\dfrac{n}{2^n}-\sum_{n=1}^{\infty}\dfrac{n}{2^n} =\sum_{n=1}^{\infty}\dfrac{n}{2^{n-1}}-\sum_{n=1}^{\infty}\dfrac{n}{2^n} =\sum_{n=0}^{\infty}\dfrac{1}{2^n}=2 \\ \end{align} $$