(-1)^n/(n) find the sum from n=0 to n=infinity

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Homework Statement



(-1)^n/(n)! find the sum from n=0 to n=infinity and do the same for (-1)^n/(2n)! Show all work

Homework Equations





The Attempt at a Solution



First answer is .367876. I just don't know how to expand out (2n)!. I know (n)! is expanded like: 1*1, 1*2, 1*2*3, 1*2*3*4... and etc. However I don't know how the expanded version of (2n)! works.
 
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You have likely derived Taylor Expansions for some common functions, like the exponential and trigonometric functions. Can you use these to evaluate your sum?
 


It looks like (-1)n/(n)! should give 1/e = e-1 ≈0.36787944117144233

Write the first several terms for each series. Follow lineintegral1's suggestions

Also, (2n)! = 1*2*3*4* ... *(n-2*(n-1)*(n)*(n+1)*(n+2)* ... *(2n-2)*(2n-1)*(2n).

(2*0)! = 0! = 1

(2*1)! = 2! = 2

(2*2)! = 4! = 24

(2*3)! = 6! = 720

...
 
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Thank you so much!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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