Monthly Archives: June 2020

Let $a_n=\sum_{d|n}\dfrac{1}{2^{d+\frac{n}{d}}}$. In other words, $a_n$ is the sum of $\dfrac{1}{2^{d+\frac{n}{d}}}$ over all divisers $d$ of $n$. Find $$\dfrac{\sum_{k=1}^{\infty}ka_k}{\sum_{k=1}^{\infty}a_k}=\dfrac{a_1+2a_2+3a_3+…}{a_1+a_2+a_3+…}$$ Click here for the solution. 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}}} \\ & = … Continue reading →