$$\sum_{n=1}^{\infty} \frac{n^{3}+\sin(-2011n)}{n^{3} + 110111}$$

I split the series

so,

sin(-2011n)/(n^(3)+110111)

Question

$$\sum_{n=1}^{\infty} \frac{n^{3}+\sin(-2011n)}{n^{3} + 110111}$$


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?

australopithecus · Answer

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

australopithecus · Answer

did I apply the comparison test correctly though?

australopithecus · Answer

sorry I'm just not confident yet with series

kinggeorge · Answer

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

australopithecus · Answer

so it is divergent?

australopithecus · Answer

can you explain why?

australopithecus · Answer

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

kinggeorge · Answer

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.

australopithecus · Answer

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

kinggeorge · Answer

Because it's just $1+1+1+1+.....$ which is infinity.

australopithecus · Answer

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

kinggeorge · Answer

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

australopithecus · Answer

thank you for clarifying :)

kinggeorge · Answer

You're welcome.