Answer
$\dfrac{3}{e(1-e)}$
Work Step by Step
Re-write the given infinite series $\Sigma_{k=2}^{\infty} 3 e^{-k}$ as $\ 3 \
\Sigma_{k=2}^{\infty} e^{-k}$
Here, we can see that the series $\ 3 \
\Sigma_{k=2}^{\infty} e^{-k}$ shows a geometric series with common ratio $r=\dfrac{1}{12}$ and $r=\dfrac{1}{e} \lt 1$. So, the series is convergent.
whose sum can be computed as: $\Sigma_{n=1}^{\infty} ar^n =a+ar+ar^2+ar^3+.....=\dfrac{a}{1-r}$
Now, $\ 3 \ \Sigma_{k=2}^{\infty} e^{-k}=3 \times \dfrac{\dfrac{1}{e^2}}{1-\dfrac{1}{e}}\\=3 \times \dfrac{1/e^2}{\dfrac{e-1}{e}}\\=\dfrac{3}{e(1-e)}$
