1. ## Test for convergence

Hey which test should I use to determine if this series converges? I'm thinking maybe a comparison test?

$\displaystyle \sum_{n=1}^\infty e^\frac{1}{n^9}$/$\displaystyle n^{10}$

2. Originally Posted by mmattson07
Hey which test should I use to determine if this series converges? I'm thinking maybe a comparison test?

$\displaystyle \sum_{n=1}^\infty e^\frac{1}{n^9}$/$\displaystyle n^{10}$
Write back because it's hard to decide what you meant to write.

Tonio

3. My apologies. I couldn't get the /frac to work around the whole expression. Its supposed to be e^(1/n^9) / n^10.

4. Originally Posted by mmattson07
Hey which test should I use to determine if this series converges? I'm thinking maybe a comparison test?

$\displaystyle \sum_{n=1}^\infty e^\frac{1}{n^9}$/$\displaystyle n^{10}$
Perhaps an integral comparison test ....

5. So since it is positive and decreasing the Integral test applies?

6. Originally Posted by mmattson07
My apologies. I couldn't get the /frac to work around the whole expression. Its supposed to be e^(1/n^9) / n^10.

It's not hard to check that $\displaystyle e^{\frac{1}{n^9}}\xrightarrow [n\to \infty] {} 1$, so you can use the comparison test for positives series: for n big enough, $\displaystyle \frac{e^{\frac{1}{n^9}}}{n^{10}}\leq \frac{2}{n^{10}}$...

Tonio