# Test this Series for Convergence or Divergence

• April 8th 2008, 05:03 PM
thegame189
Test this Series for Convergence or Divergence
Test this Series for Convergence or Divergence.

Summation from 1 to infinity of (-1)^n * sin(8Pi/n)

It's from the Alternating Series Test section. I wasn't sure how to do this. I compared b(sub n)= sin(8Pi/n) to b(sub n+1)=sin(8Pi/n+1), b(sub n+1) is always less than b(sub n), so I just take the limit of b(sub n), right? If it's 0, it's convergent?

I don't have my book with me so I'm not quite sure.
• April 8th 2008, 05:21 PM
Mathstud28
Here you go
[quote=thegame189;126859]Test this Series for Convergence or Divergence.

Summation from 1 to infinity of (-1)^n * sin(8Pi/n)

It's from the Alternating Series Test section. I wasn't sure how to do this. I compared b(sub n)= sin(8Pi/n) to b(sub n+1)=sin(8Pi/n+1), b(sub n+1) is always less than b(sub n), so I just take the limit of b(sub n), right? If it's 0, it's convergent?

you have $\sum_{n=1}^{\infty}\bigg[sin\bigg(\frac{8\pi}{n}\bigg)\cdot{cos(\pi{n})}\bi gg]$...does that make it clearer?
• April 8th 2008, 05:26 PM
thegame189
Not at all (Worried)

Where did the cos(nPi) come from?
• April 8th 2008, 05:33 PM
Mathstud28
Ok
Quote:

Originally Posted by thegame189
Not at all (Worried)

Where did the cos(nPi) come from?

It came from the fact that for integer values for x... $(-1)^{x}=cos{\pi{x}}$..think about it $cos(\pi{\cdot{1}})=-1,cos(\pi\cdot{2})=-1,cos(\pi\cdot{3})=-1...$ ad infinitum...so is it easier to work with now?
• April 8th 2008, 05:47 PM
PaulRS
Since it's a alternating series, $\lim_{n\rightarrow{+\infty}}{\sin\left(\frac{8\cdo t{\pi}}{n}\right)}=0$ *, and $\sin\left(\frac{8\cdot{\pi}}{n}\right)$ is decreasing, it converges

Alternating series test - Wikipedia, the free encyclopedia

* $\sin(x)\approx{x}$ as $x\rightarrow{0}$