# Series - comparison test

1. Oct 20, 2009

### IniquiTrance

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

I need to show that:

$$\sum_{n=1}^\infty \frac{n^{2}}{2^n}$$

converges. I know I can compare it with the larger convergent geometric series:

$$\sum_{n=1}^\infty \frac{1.5^{n}}{2^n}$$

Which is larger for all terms for n> 13.

My question is, I found this "13" through trial and error. Is there any concrete way of determining when bn becomes larger than nc?

Thanks!
2. Relevant equations

3. The attempt at a solution

2. Oct 20, 2009

### Bohrok

You could use induction to show that it works for all n>13. However, using the ratio test or root test would be much easier for this series.

3. Oct 20, 2009

### IniquiTrance

Yeah the problem actually asks me to put an upper bound on Sn, so I needed to compare it to a geometric series with a known sum. So trial and error is the only way to go to get that 13?

4. Oct 20, 2009

### Bohrok

Finding that 13 by trial and error isn't important; what is important is that you can show that it's true for n>13.

5. Oct 20, 2009

### IniquiTrance

But I need to know at which point it becomes true to collect a partial sum...

6. Oct 21, 2009

### IniquiTrance

Ok, so basically the problem asks me to place an upper bound on:

$$\sum_{n=1}^\infty \frac{n^{2}}{2^n}$$

So what I did was use this:

$$\sum_{n=1}^{13} \frac{n^{2}}{2^n} + \sum_{n=13}^\infty \frac{1.5^{n}}{2^n}$$

as my bound.

Is there a more methodical way than trial and error to figure out at which point the geomtric series is larger than the original one?