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Mathematics 17 Online
OpenStudy (anonymous):

Have can I show this series are convergent or divergent?

OpenStudy (anonymous):

OpenStudy (anonymous):

Use the squeeze theorem and comparison tests

OpenStudy (chillout):

Wouldn't it be Leibniz's test?

OpenStudy (anonymous):

I have never head about the squeeze theorem and neither the Leibniz's test. Can someone see another way to solve the series?

OpenStudy (anonymous):

of course I know the comparison test..

OpenStudy (zarkon):

just use the limit comparison test

OpenStudy (chillout):

Leibniz's test is the one for alternating series. Squeeze theorem is the one we compare with two other functions' limits, generally easier ones. And yeah, we don't use Leibniz's test here.

OpenStudy (anonymous):

the series shall not be positive to use the limit comparison test?

OpenStudy (zarkon):

again, just use the limit comparison test

OpenStudy (zarkon):

\[\frac{\frac{n+(-1)^n7}{2n^2+1}}{\frac{1}{n}}\to\frac{1}{2}\text{ as }n\to\infty\]

OpenStudy (anonymous):

yes, but the series should not be positive in order to use the limit comparison?

OpenStudy (anonymous):

Or are I missing something?

OpenStudy (zarkon):

it only matters that the series is eventually positive

OpenStudy (chillout):

just the simple comparison one. But isn't 1/2 inconclusive?

OpenStudy (zarkon):

no...it tells us that the original series diverges

OpenStudy (chillout):

I mean, 0<=limt<=inf both can diverge and both can converge.

OpenStudy (zarkon):

\[0<1/2<\infty\] and \[\sum_{n=1}^{\infty}\frac{1}{n}\] diverges

OpenStudy (chillout):

Oh, right... I see. I'm also learning series.

OpenStudy (chillout):

I just didn't see the previous posts. That's why I missed 1/n :P

OpenStudy (anonymous):

Thank you @Zarkon

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