## agentx5 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

1. agentx5

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

2. agentx5

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

3. mukushla

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

4. agentx5

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

5. mukushla

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6. FoolAroundMath

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)$$

7. mukushla
8. agentx5

Awesome thanks for your help! :-)