HMMT-Feb 2010 calc, Problem 8

Let $f(n)=\sum_{k=2}^{\infty} \frac{1}{k^{n} \cdot k !}$. Calculate $\sum_{n=2}^{\infty} f(n)$.

Solution: \[ \begin{aligned} \sum_{n=2}^{\infty} f(n) & =\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{k^{n} \cdot k !} \\ & =\sum_{k=2}^{\infty} \frac{1}{k !} \sum_{n=2}^{\infty} \frac{1}{k^{n}} \\ & =\sum_{k=2}^{\infty} \frac{1}{k !} \cdot \frac{1}{k(k-1)} \\ & =\sum_{k=2}^{\infty} \frac{1}{(k-1) !} \cdot \frac{1}{k^{2}(k-1)} \end{aligned} \] \[ \begin{aligned} & =\sum_{k=2}^{\infty} \frac{1}{(k-1) !}\left(\frac{1}{k-1}-\frac{1}{k^{2}}-\frac{1}{k}\right) \\ & =\sum_{k=2}^{\infty}\left(\frac{1}{(k-1)(k-1) !}-\frac{1}{k \cdot k !}-\frac{1}{k !}\right) \\ & =\sum_{k=2}^{\infty}\left(\frac{1}{(k-1)(k-1) !}-\frac{1}{k \cdot k !}\right)-\sum_{k=2}^{\infty} \frac{1}{k !} \\ & =\frac{1}{1 \cdot 1 !}-\left(e-\frac{1}{0 !}-\frac{1}{1 !}\right) \\ & =3-e \end{aligned} \] $\fbox{3-e}$.

Answer: 3-e