# Math Help - series

1. ## series

test for convergence/divergence

$\sum_{n =1}^{\infty}\frac{n^2+1}{5^n}$

2. $\sum_{n=0}^{\infty} \frac{n^2+1}{5^n} = \sum_{n=0}^{\infty} \frac{n^2}{5^n} + \sum_{n=0}^{\infty} \frac{1}{5^n}$

If the two series on the right are convergent, then so is the one on the left. The right-most series is clearly geometric, so it's just the left series on the RHS we have to deal with.

If you know the ratio test, you can use it to find that $\sum_{n = 0}^{\infty} \frac{n^2}{5^n}$ converges.

If you haven't learned it yet, use the identity: $n^2 < 2^n$ for all $n \geq 5$ and use comparison.

3. i'm just wondering..does it make any difference that the series that i posted started at n=1 and you're using n=0?

also, is there anyway that you could kind of show me how the ratio test works here? i just learned it and i'm a little unsure of myself

4. Originally Posted by buttonbear
i'm just wondering..does it make any difference that the series that i posted started at n=1 and you're using n=0?

Mr F says: No. That is simply a typo by o_O.

also, is there anyway that you could kind of show me how the ratio test works here? i just learned it and i'm a little unsure of myself
You should first review the examples in your textbook and/or classnotes. Also, you should review the many examples to be found on the world wide web. eg. Pauls Online Notes : Calculus II - Ratio Test

5. i just tried to use the ratio test on this and i ended up getting that the limit of $\frac{a(n+1)}{an}={\infty}$ where $an= \frac{n^2}{5^n}$ which means it diverges, but someone on this thread said it converges..so i must be doing something wrong..please help?

6. Originally Posted by buttonbear
i just tried to use the ratio test on this and i ended up getting that the limit of $\frac{a(n+1)}{an}={\infty}$ where $an= \frac{n^2}{5^n}$ which means it diverges, but someone on this thread said it converges..so i must be doing something wrong..please help?
$a_n = \frac{n^2}{5^n}$.

$\left| \frac{a_{n+1}}{a_n} \right| = \frac{ \frac{(n+1)^2}{5^{n+1}} }{\frac{n^2}{5^n}} = \frac{5^n}{5^{n+1}} \cdot \frac{(n+1)^2}{n^2}$.

Simplify this expression and take the limit $n \rightarrow +\infty$. The limiting value is $\frac{1}{5}$.