Answer
Converges to $\frac{1}{2}e^{-2}$
Work Step by Step
Consider $f(x)=e^{-2x}$. This is positive, continuous and decreasing for $x \geq 1$
Now, take the integral test to find the convergence and divergence for the sequence.
We have $\int_1^\infty e^{-2x} dx= \lim\limits_{a \to \infty} \int_1^a e^{-2x} dx$
or, $ \lim\limits_{a \to \infty} [\dfrac{-1}{2} e^{-2x}]_1^a=\lim\limits_{a \to \infty} [\dfrac{-1}{2} e^{-2a}+\frac{1}{2}e^{-2}]= \frac{1}{2}e^{-2}$
Hence, the sequence $\Sigma_{n=1}^\infty e^{-2n}$ converges to $\frac{1}{2}e^{-2}$
