Thread: infinite series

test for convergence/divergence of the series $\displaystyle \sum_{n =1}^{\infty}\frac{e^{1/n}}{n^2}$

i'm just wondering if i should use the integral test here..or if it can just be compared to the p series (p=2)

Originally Posted by buttonbear

test for convergence/divergence of the series $\displaystyle \sum_{n =1}^{\infty}\frac{e^{1/n}}{n^2}$

i'm just wondering if i should use the integral test here..or if it can just be compared to the p series (p=2)

Direct comp, since $\displaystyle e^{1/n}\to 1 $ as $\displaystyle n\to\infty$

use $\displaystyle 0<e^{1/n}<100$ so $\displaystyle \sum_{n =1}^{\infty}\frac{e^{1/n}}{n^2} < 100\sum_{n =1}^{\infty}\frac{1}{n^2}<\infty$

where the 100 is larger than what you need, but it get the point across.
The max of $\displaystyle e^{1/n}$ is actually $\displaystyle e$. So, any number $\displaystyle e$ or larger works, but that's not important.

Originally Posted by buttonbear

test for convergence/divergence of the series $\displaystyle \sum_{n =1}^{\infty}\frac{e^{1/n}}{n^2}$

i'm just wondering if i should use the integral test here..or if it can just be compared to the p series (p=2)

To answer your question, yes the integral test can be used and also a limit comparision with $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$. It might be a bit easier the a direct comparision test.

okay, so, just to be clear..it CAN be compared to the p-series? i'm just not 100% sure that i understand all of that..and i need to be able to explain my work. thanks again!

Yes, I did a direct comp test to the p=2 series.
You can do a limit comp test here too, with p=2 also.

thanks so much