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

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Homework Help Overview

The problem involves finding the sum of the series \((-1)^n/(n)!\) and \((-1)^n/(2n)!\) from \(n=0\) to \(n=\infty\). The original poster expresses uncertainty about expanding \((2n)!\) and seeks clarification on the series.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of Taylor expansions for common functions to evaluate the sums. The original poster attempts to understand the expansion of \((2n)!\) and how it relates to the series.

Discussion Status

Some participants have provided insights into the evaluation of the series and suggested writing out the first several terms. There is an acknowledgment of the original poster's confusion regarding the factorial expansion.

Contextual Notes

The original poster has provided an initial numerical result for the first series but is unsure about the factorial expansion for the second series. There is a lack of consensus on the approach to take for evaluating \((-1)^n/(2n)!\).

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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

...
 
Last edited:


Thank you so much!
 

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