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

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agentx5Best ResponseYou've already chosen the best response.0
I got to go for now but I'll be back tomorrow to see if anybody has figured this one out :)
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Convergence/Divergence and if it converges to what are the typical questions for these...
 one year ago

mukushlaBest ResponseYou've already chosen the best response.5
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 \(\large \sum_{n=1}^{\infty} \frac{1}{n!}=e1\)  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
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
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
 one year ago

mukushlaBest ResponseYou've already chosen the best response.5
dw:1342891305015:dw
 one year ago

FoolAroundMathBest ResponseYou've already chosen the best response.0
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! + ... = \((e^{i}  e^{i})/2i = sin(1)\)
 one year ago

mukushlaBest ResponseYou've already chosen the best response.5
@agentx5 http://www.sosmath.com/calculus/series/absolute/absolute.html
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Awesome thanks for your help! :)
 one year ago
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