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Mathematics 15 Online
OpenStudy (australopithecus):

\[\sum_{n=1}^{\infty} \frac{n^{3}+\sin(-2011n)}{n^{3} + 110111}\] I split the series so, sin(-2011n)/(n^(3)+110111) < or = 1/n^(3) by the p test the series converges and by comparison the original series also converges then for the other series n^(3)/(n^(3) + 110111) < or = 1 My main question does a series with a constant converge? I would think it would?

OpenStudy (australopithecus):

my question is \[\sum_{n=1}^{\infty} 1\] =1 thus the series converges

OpenStudy (australopithecus):

did I apply the comparison test correctly though?

OpenStudy (australopithecus):

sorry I'm just not confident yet with series

OpenStudy (kinggeorge):

Sorry, afk for a second there. But \[\sum_{n=1}^\infty 1 \neq1\]

OpenStudy (australopithecus):

so it is divergent?

OpenStudy (australopithecus):

can you explain why?

OpenStudy (australopithecus):

If one of the series is divergent the entire series is divergent

OpenStudy (kinggeorge):

Correct. So you split it up into \[\sum_{n=1}^\infty \frac{n^3}{n^3+110111} +\sum_{n=1}^\infty \frac{\sin(-2011n)}{n^3+110111}\]Since the first part is divergent, the whole thing is divergent.

OpenStudy (australopithecus):

yes but why does \[\sum_{n=1}^{\infty} 1\] diverge?

OpenStudy (kinggeorge):

Because it's just \(1+1+1+1+.....\) which is infinity.

OpenStudy (australopithecus):

yeah that makes sense I see that the n=1 is not applicable when there is a constant now

OpenStudy (kinggeorge):

You can also use a comparison test on \[\sum_{n=1}^\infty \frac{1}{n}\]

OpenStudy (australopithecus):

thank you for clarifying :)

OpenStudy (kinggeorge):

You're welcome.

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