How do I determine if the series $ e^{-2n} $ converges or diverges? (From $ 0\to \infty $)

Question

I need homework help, I am mostly getting hung up on the integral. The question is:

Determine if the series $$\sum_{n=1}^{\infty}{e^{-2n}}$$ converges or diverges.

Again, it isn't the concept I don't understand, it is the integral of e. Anything helps, thank you!

Here's what I have thus far:

$$\int_1^\infty e^{-2x}dx=\lim_{b\to\infty}\left[-\frac{e^{-2x}}{2}\right]_1^b=\lim_{b\to\infty}\left(-\frac{e^{-2b}}{2}\right)-\left(-\frac{e^{-2}}{2}\right)$$

Answer 1

Hint:

This is a geometric sum, think of when does geometric sum converges? what is the common ratio here?

Edit:

You actually almost did it.

$$\lim_{b \to \infty} \left(-\frac{e^{-2b}}{2} \right)-\left(-\frac{e^{-2}}{2} \right)=\frac{e^{-2}}{2}<\infty, $$

Hence it converges.

Answer 2

By the integral test, $\sum_{n=0}^\infty f(n) < \infty$ iff $\int_0^\infty f(x) dx < \infty$. Can you test it?

UPDATE

Use $u = -2x$ then $du = -2dx$ and $$ \int_0^\infty e^{-2x} dx = \int_{u=0}^{u=-\infty} e^u \frac{du}{-2} = \frac{1}{-2} \int_0^{-\infty}e^u du. $$ Can you finish this now?

Answer 3

Your series is a geometric series, so you can just look up the partial sums:

$$\sum_{n=0}^N e^{-2n}=\frac{1-e^{-2(N+1)}}{1-e^{-2}}\xrightarrow[N\to\infty]{} \frac 1{1-e^{-2}}.$$

So your series converges (and you even have the value it converges to !).

Answer 4

See here: https://en.wikipedia.org/wiki/Integral_test_for_convergence

So you need to integrate this: $$\int_{1}^{\infty}{e^{-2n}}dn$$$$\rightarrow \bigg[-\frac{1}{2}e^{-2n}\bigg]^{\infty}_{1}$$ $$\rightarrow-\frac{1}{2e^{2\infty}}-\frac{1}{2e^2}$$

To say $2e^{2\infty}$ isnt technically correct, but appreciate that $\frac{1}{2e^L}\rightarrow0$ as $L\to \infty $, hence the result is convergence to $\frac{1}{2e^2}$

Answer 5

You're almost there. That term on the right is constant, so forget about it. You have $\lim_{b\to\infty} \frac{-e^{-2b}}{2} = -\frac12\lim_{b\to\infty}e^{-2b}$. What happens if you take $e$ to the power of a large negative number? (You can think of $e^{-2b}$ as $\frac1{e^{2b}}$ instead if that helps.)