Answer
Divergent series
Work Step by Step
Let us consider a geometric series $\Sigma_{n=k}^{n=\infty} ar^n$ when $a \ne 0$. This series will diverge when $|r| \gt1$ and the sum of this series can be computed as: $S=\dfrac{ar^k}{1-r}$
We are given the series $\Sigma_{k=1}^{\infty} \dfrac{\pi^k}{e^{k+1}}=\Sigma_{k=1}^{\infty}(\dfrac{1}{e})( \dfrac{\pi}{e})^k$
Here, we have $r=\dfrac{\pi}{e} \gt 1$
So, we can conclude the sum of the given geometric series is undetermined and the series is divergent.
