# Infinite series, does ∑ n/2^n diverge?

1. Oct 20, 2009

### utleysthrow

1. The problem statement, all variables and given/known data

$$\sum^{\infty}_{n=1} \frac{n}{2^{n}}$$

Does this series converge or diverge?

2. Relevant equations

3. The attempt at a solution

By the Cauchy condensation test (http://en.wikipedia.org/wiki/Cauchy_condensation_test) I think this one diverges. But not sure if I am using it correctly.

According to the test,

$$\sum^{\infty}_{n=1} \frac{n}{2^{n}}$$

converges if and only if

$$\sum^{\infty}_{n=1} 2^{n} \frac{2^{n}}{2^{n}} = \sum^{\infty}_{n=1} 2^{n}$$

converges, which doesn't.

Thank you for any help.

2. Oct 20, 2009

### Staff: Mentor

In the line above, you aren't using the condensation test correctly. For your series, f(n) = n/(2n), so what would be f(2n)?

A test that would be simpler to apply would be the Limit Ratio Test.

3. Oct 20, 2009

### utleysthrow

Ah, okay, I see it where I went wrong..

Using the limit ratio test

$$lim \left| a_{n+1}/a_{n} \right| = lim \left| \frac{(n+1)/2^{n+1}}{n/2^{n}} \right| = 1/2 < 1$$

So it converges...

4. Oct 21, 2009

Yes, indeed.