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

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In summary, the conversation discusses finding the sum of (-1)^n/(n)! and (-1)^n/(2n)! from n=0 to n=infinity. The first series evaluates to approximately 0.367876, while the second series can be expanded using Taylor Expansions and evaluates to e-1 ≈ 0.36787944117144233. The conversation also provides the first few terms of the expanded version of (2n)!.
  • #1
chupe
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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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  • #2


You have likely derived Taylor Expansions for some common functions, like the exponential and trigonometric functions. Can you use these to evaluate your sum?
 
  • #3


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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  • #4


Thank you so much!
 

1. What is the formula for calculating the sum of (-1)^n/(n) from n=0 to n=infinity?

The formula for calculating the sum of (-1)^n/(n) from n=0 to n=infinity is (-1)^n/(n) = ln(2) .

2. Is the sum of (-1)^n/(n) from n=0 to n=infinity a convergent or divergent series?

This series is convergent as it approaches a finite value (ln(2)) as n approaches infinity.

3. What is the value of the sum of (-1)^n/(n) from n=0 to n=infinity?

The value of the sum of (-1)^n/(n) from n=0 to n=infinity is ln(2) .

4. Can the sum of (-1)^n/(n) from n=0 to n=infinity be approximated?

Yes, the sum can be approximated by summing a finite number of terms in the series. The more terms that are used, the closer the approximation will be to the actual value of ln(2).

5. How is the sum of (-1)^n/(n) from n=0 to n=infinity related to the natural logarithm function?

The sum of (-1)^n/(n) from n=0 to n=infinity is equal to the natural logarithm of 2 (ln(2)). This relationship can be seen in the formula for calculating the sum and the properties of logarithms.

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