Determine whether the series converges or diverges.
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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
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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
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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.