Answer
converges to $2e$
Work Step by Step
Let us consider $f(x)=e^{-2x}$.
Here, the function $f(x)$ is positive, continuous and decreasing for $x \geq 1$
Then $\int_1^\infty e^{-2x} dx= \lim\limits_{k \to \infty} \int_1^k e^{-2x} dx=\lim\limits_{a \to \infty} [\dfrac{-1}{2} e^{-2x}]_1^k$
and $\lim\limits_{k \to \infty} [\dfrac{-1}{2} e^{-2k}+2e]= 2e$
Thus, the sequence $\Sigma_{n=1}^\infty e^{-2n}$ converges to $2e$.
