Finding the Upper Bound for a Convergent Series: A Comparison Method

[IniquiTrance]

Homework Statement

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 n^c?

Thanks!

Homework Equations

The Attempt at a Solution

 

[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.

 

[IniquiTrance]

Yeah the problem actually asks me to put an upper bound on S_n, 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?

 

[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.

 

[IniquiTrance]

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

 

[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?