Use Integral Test to prove convergence/divergence

Question

$$\sum_{n=0}^\ \frac{n}{e^{2n}}$$

Use the Integral Test to prove convergence/divergence

Ok I know the integral test is:

1. $f(x)$ is supposed to be continuous
2. $f(x)$ is decreasing
3. $f(x)$ is positive

And I understand how to format the integral. the problem is integrating that function. I've tried everything but i am not sure that it is integrable

Answer 1

You don't need to integrate it. Just use the fact that$$\frac x{e^{2x}}\leqslant\frac{e^x}{e^{2x}}=e^{-x}$$and that the integral $\displaystyle\int_0^\infty e^{-x}\,\mathrm dx$ converges.

Answer 2

$$\int xe^{-2x}dx=-\frac12 xe^{-2x}-\frac14 e^{-2x}+C$$ By applying integration by parts with $u=x$ and $\frac{dv}{dx}=e^{-2x}$.

Answer 3

Hint: $$\frac{n}{e^{2n}}=ne^{-2n}$$ which means you can write your integrand as $xe^{-2x}$. Do you recognize which technique to use to integrate this function?

Answer 4

Hint: $$\sum_{n=0}^{\infty}nx^{2n}=\frac{x^2}{(1-x^2)^2}$$