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Mathematics 22 Online
OpenStudy (anonymous):

Determine whether the series converges or diverges.

OpenStudy (anonymous):

\[\sum_{n=1}^{\infty}\frac{ 1 }{ 2n-1 }\]

OpenStudy (anonymous):

the values that are added on top get smaller and smaller when they get close to infinity, they contribute almost nothing ( 0) so I think the series converges to a value

OpenStudy (anonymous):

how would i prove that? is this a p-series?

OpenStudy (anonymous):

would i use the integral test?

OpenStudy (anonymous):

Integral test would work

OpenStudy (anonymous):

how would i set that up? im used to doing it with only one numberin the numerator.

OpenStudy (anonymous):

denominator*

OpenStudy (anonymous):

i got it! i got the answer to =1/

OpenStudy (anonymous):

Basically, \[\sum_{n=1}^{\infty} \left(\frac{1}{2n - 1}\right)\] converges if and only if \[\int_1^\infty \frac{1}{2x - 1} \mathrm{d}x\] is finite

OpenStudy (anonymous):

The integral itself diverges (you can check this yourself probably), so the series diverge aswell

OpenStudy (anonymous):

okay what about \[\frac{ 1 }{ n \sqrt{n^{2}-1} }\]

OpenStudy (anonymous):

From n = 1?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

There's no sum if you start from n = 1, since when n = 1 you're dividing by zero

OpenStudy (anonymous):

thts what i had so i just said it diverged becasuse the limit does not exist.

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