# Testing a series for convergence or divergence

• Apr 22nd 2010, 09:21 PM
CalculusCrazed
Testing a series for convergence or divergence
I am trying to test this series by the alternating series test:

$\displaystyle \sum\frac{\sin{n}\frac\pi2}{n!}$

n=1 to infinity

I don't really know how to find the limit or how to see if it is decreasing, any help would be greatly appreciated.

Thanks,
CC
• Apr 22nd 2010, 09:58 PM
Drexel28
Quote:

Originally Posted by CalculusCrazed
I am trying to test this series by the alternating series test:

$\displaystyle \sum\frac{\sin{n}\frac\pi2}{n!}$

n=1 to infinity

I don't really know how to find the limit or how to see if it is decreasing, any help would be greatly appreciated.

Thanks,
CC

$\displaystyle \sin\left(\frac{n\pi}{2}\right)=(-1)^n$
• Apr 22nd 2010, 10:08 PM
CalculusCrazed
So I basically have:

$\displaystyle \lim\frac{1}{n!}$
as n goes to infinity

which would equal zero

and decreasing because the first term is
1/1 then the second is 1/(2*1) and so on?
• Apr 22nd 2010, 10:11 PM
Drexel28
Quote:

Originally Posted by CalculusCrazed
So I basically have:

$\displaystyle \lim\frac{1}{n!}$
as n goes to infinity

which would equal zero

and decreasing because the first term is
1/1 then the second is 1/(2*1) and so on?

Oh wow, I didn't even notice that it was $\displaystyle n!$.

With my hint you can use the alternating series test.

You can also note that $\displaystyle \left|\sum_{n=1}^{\infty}\frac{\sin\left(\frac{\pi n}{2}\right)}{n!}\right|\leqslant$$\displaystyle \sum_{n=1}^{\infty}\frac{\left|\sin\left(\tfrac{\p i n}{2}\right)\right|}{n!}$$\displaystyle \leqslant\sum_{n=1}^{\infty}\frac{1}{n!}=e$

Or the ratio test even works nice.
• Apr 22nd 2010, 10:18 PM
CalculusCrazed
Ahh, okay cool. Thanks a ton!