Question

Ok honestly the notes make no sense on this one... How do I do this?

\[\huge \sum_{n=1}^{\infty} \frac{\sin(\frac{\pi n}{2})}{n!}\]

Answer 1

I got to go for now but I'll be back tomorrow to see if anybody has figured this one out :-)

Answer 2

Convergence/Divergence and if it converges to what are the typical questions for these...

Answer 3

hi agentx5

1.Convergence/Divergence

u can write \( \large 0 \le |\frac{\sin \frac{n \pi}{2}}{n!}| \le \frac{1}{n!} \) and use this \(\large \sum_{n=1}^{\infty} \frac{1}{n!}=e-1\)
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2.Convergence to What

see \(\large \sin \frac{n \pi}{2}=0 \ \ for \ \ n=2,4,6,... \) and

\(\large \sin \frac{n \pi}{2}=1 \ or \ -1 \ \ for \ \ n=1,3,5,... \) so u can rearrange the sigma like this

\[\large \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\]
now think of Maclaurin series for sin to Evaluate the sum

Answer 4

So kind of like the Squeeze Theorem for the first one, gotcha. Does this test have a name though? Typically we have to say something like "convergent by the alternating series test" or something like that.

I don't think we've gotten to MacLaurin series yet, but I'll try to understand :-)
Would #46 from this be what I'm looking for? http://mathworld.wolfram.com/MaclaurinSeries.html