$\sum_{m=1}^{\infty} \frac{1}{m!}\left(\sum_{n=1}^m 3^{n-1}\right)$ is equal to

Question

Professor PC Thomas Classes · Answer

Inner sum is $\frac{3^m-1}{2}$. Series becomes $\frac{1}{2} \sum (\frac{3^m}{m!} - \frac{1}{m!}) = \frac{1}{2}(e^3 - e)$.