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anonymous
 4 years ago
Ok honestly the notes make no sense on this one... How do I do this?
\[\huge \sum_{n=1}^{\infty} \frac{\sin(\frac{\pi n}{2})}{n!}\]
anonymous
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got to go for now but I'll be back tomorrow to see if anybody has figured this one out :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Convergence/Divergence and if it converges to what are the typical questions for these...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hi agentx5 1.Convergence/Divergence u can write \( \large 0 \le \frac{\sin \frac{n \pi}{2}}{n!} \le \frac{1}{n!} \) and use this \(\large \sum_{n=1}^{\infty} \frac{1}{n!}=e1\)  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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342891305015:dw

FoolAroundMath
 4 years ago
Best ResponseYou've already chosen the best response.0Using 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! + ... = \((e^{i}  e^{i})/2i = sin(1)\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@agentx5 http://www.sosmath.com/calculus/series/absolute/absolute.html

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Awesome thanks for your help! :)
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