Question

Determine convergence or diverge of ((-1)^n)((2)^n)/(n^3)

Answer 1

Please someone help. This is bugging me and I can't sleep.

Answer 2

The -(1)^n doesn't matter much. It just lets the terms alternate between - and +.
Consider the function \[f(x)=\frac{ 2^x }{ x^3 }\]Because \[\lim_{x \rightarrow \infty}2^x=\infty \]and\[\lim_{x \rightarrow \infty}x^3=\infty \]you could use l'Hôpital's Rule (three times after another, actually) to finally get\[\lim_{x \rightarrow \infty}\frac{ \ln ^3 2}{ 6 }2^x=\frac{ \ln ^3 2}{ 6 }\lim_{x \rightarrow \infty}2^x=\frac{ \ln ^3 2}{ 6 }*\infty=\infty\]
So I would say that this sequence diverges.

On the other hand, you might not know l'H, or want to make a more sophisticated etimation of the values of 2^n and n^3 to see which of the two will "win"...
Anyway, I hope you will be able to sleep now ;)

Answer 3

Thank you! i actually had to do the root test to find that it converges.

Answer 4

Could you show me your calculations?
I think this sequence is divergent.

Answer 5

\[\sum_{1}^{\infty} \frac{ (-1)^n(2)^n }{ n^3 } = \sum_{1}^{\infty} \frac{ (-2)^n }{ n^3 }\]

Answer 6

From there I did the Ratio Test and got a convergence.

Answer 7

WolframAlpha thinks it is not convergent...
http://www.wolframalpha.com/input/?i=sum+%28%28-1%29^n%29%28%282%29^n%29%2F%28n^3%29%2C+n%3D+1+to+infinity

Answer 8

My professor was wrong! I'll show him this. Thank you!

Answer 9

YW!