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Ok honestly the notes make no sense on this one... How do I do this? \[\huge \sum_{n=1}^{\infty} \frac{\sin(\frac{\pi n}{2})}{n!}\]

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I got to go for now but I'll be back tomorrow to see if anybody has figured this one out :-)
Convergence/Divergence and if it converges to what are the typical questions for these...
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 \(\large \sum_{n=1}^{\infty} \frac{1}{n!}=e-1\) ------------------------------------------------------------------------- 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

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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?
Using the power series expansion for e^(x) around x=0 (which is equivalent to Maclaurin's series) \[e^{i} = 1 + i - 1/2! -i/3! + 1/4! + i/5! - 1/6! - i/7! + ....\]\[e^{-i} = 1 - i - 1/2! + i/3! + 1/4! - i/5! - 1/6! + i/7! + ...\] what we need is 1 - 1/3! + 1/5! - 1/7! + ... try subtracting second series from first to get \[e^{i} - e^{-i} = 2i(1 - 1/3! + 1/5! - 1/7! + ... )\]So, 1 - 1/3! + 1/5! - 1/7! + ... = \((e^{i} - e^{-i})/2i = sin(1)\)
Awesome thanks for your help! :-)

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