Answer
$\dfrac{2}{e-2}$
Work Step by Step
A geometric series $\Sigma_{k=0}^{\infty} ar^k$ is said to be converges when $|r| \lt 1 $ and diverges when $|r| \geq 1$
and its sum is given by: $\Sigma_{k=0}^{\infty} ar^k=\dfrac{a}{1-r}$
Re-write the given series as: $\Sigma_{k=1}^{\infty} (\dfrac{2}{e})^{k}= \dfrac{2}{e}+\dfrac{2}{e} \cdot \dfrac{2}{e} +\dfrac{2}{e} \cdot (\dfrac{2}{e})^2+...... $
We find that $r=\dfrac{2}{e} \lt 1$.
So, its sum can be computed as:
$\Sigma_{k=1}^{\infty} (\dfrac{2}{e})^{k}=\dfrac{2/e}{1-\dfrac{2}{e}}\\=\dfrac{2/e}{\dfrac{e-2}{e}}\\=\dfrac{2}{e-2}$
