Ratio Test problem!!

ecwang

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May 5, 2021
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The problem is the series of (n!(-1)^(n+1))/(1 · 3 · 5 · ... (2n+1)) from n=0 to infinity.

According to my textbook, the answer to the limit is 1/2 and it absolutely converges. I am stuck on solving the limit! Any help is appreciated, thank you!! <3
 
I’m sorry, but I cant figure out the expression. Please write it in LaTeX or by hand, and upload the image here.
 
The problem is the series of (n!(-1)^(n+1))/(1 · 3 · 5 · ... (2n+1)) from n=0 to infinity.

According to my textbook, the answer to the limit is 1/2 and it absolutely converges. I am stuck on solving the limit! Any help is appreciated, thank you!! <3
Is your problem:

0(1)n+1n!1357...(2n+1)\displaystyle \sum_0^\infty \frac {(-1)^{n+1}*n!}{1*3*5*7... * (2n+1)} ................. If it is then convert it to:

0(1)n+1[2468....(2n)]n!12345678...(2n+1)\displaystyle \sum_0^\infty \frac {(-1)^{n+1} * [2*4*6*8*.... *(2n)] * n!}{1*2*3*4*5*6*7*8... * (2n+1)} ...... continue .....

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem
 
[MATH]1 \cdot 3 \cdot 5 \cdot \, ... \, \cdot (2n-1) \cdot (2n+1) = \frac{1 \cdot 2 \cdot 3 \cdot \, ... \, \cdot (2n+1)}{2(1 \cdot 2 \cdot 3 \cdot \, ... \, \cdot n)} = \frac{(2n+1)!}{2 \cdot n!}[/MATH]
[MATH]\sum_{n=0}^\infty \frac{2 \cdot n! \cdot n! \cdot (-1)^{n+1}}{(2n+1)!}[/MATH]
limn2(n+1)!(n+1)!(1)n+2(2n+3)!(2n+1)!2n!n!(1)n+1\lim_{n \to \infty} \bigg| \frac{2 \cdot (n+1)! \cdot (n+1)! \cdot (-1)^{n+2}}{(2n+3)!} \cdot \frac{(2n+1)!}{2 \cdot n! \cdot n! \cdot (-1)^{n+1}} \bigg|
The series converges, but unless I'm making an error somewhere, I do not get 1/2 as the limit.
 
[MATH]1 \cdot 3 \cdot 5 \cdot \, ... \, \cdot (2n-1) \cdot (2n+1) = \frac{1 \cdot 2 \cdot 3 \cdot \, ... \, \cdot (2n+1)}{2(1 \cdot 2 \cdot 3 \cdot \, ... \, \cdot n)} = \frac{(2n+1)!}{2 \cdot n!}[/MATH]The series converges, but unless I'm making an error somewhere, I do not get 1/2 as the limit.

[MATH]1 \cdot 3 \cdot 5 \cdot \, ... \, \cdot (2n-1) \cdot (2n+1) = \frac{1 \cdot 2 \cdot 3 \cdot \, ... \, \cdot (2n+1)}{\boldsymbol{2^{n}}(1 \cdot 2 \cdot 3 \cdot \, ... \, \cdot n)} = \frac{(2n+1)!}{\boldsymbol{2^{n}} \cdot n!}[/MATH]and then the ratio test does give a limit of 1/2.
 
[MATH]1 \cdot 3 \cdot 5 \cdot \, ... \, \cdot (2n-1) \cdot (2n+1) = \frac{1 \cdot 2 \cdot 3 \cdot \, ... \, \cdot (2n+1)}{\boldsymbol{2^{n}}(1 \cdot 2 \cdot 3 \cdot \, ... \, \cdot n)} = \frac{(2n+1)!}{\boldsymbol{2^{n}} \cdot n!}[/MATH]and then the ratio test does give a limit of 1/2.

forgot the exponent on 2 ... mea culpa. Thanks.
 
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