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

[utleysthrow]

Homework Statement

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

Does this series converge or diverge?

Homework Equations

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.

 

[Mark44]

Homework Statement

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

Does this series converge or diverge?

Homework Equations

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.

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.

Thank you for any help.

 

[utleysthrow]

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.

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...

 

[Mark44]

Yes, indeed.