infinite series

**buttonbear** · March 27th 2009, 10:01 PM

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

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)

**matheagle** · March 27th 2009, 11:50 PM

Originally Posted by buttonbear

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

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 $e^{1/n}\to 1$ as $n\to\infty$

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

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

**Danny** · March 28th 2009, 06:06 AM

Originally Posted by buttonbear

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

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 $\sum_{n=1}^\infty \frac{1}{n^2}$ . It might be a bit easier the a direct comparision test.

**buttonbear** · March 28th 2009, 07:38 PM

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!

**matheagle** · March 28th 2009, 07:57 PM

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.

**buttonbear** · March 28th 2009, 08:26 PM

thanks so much

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