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Series with Factorial 
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#1
Mar1106, 12:35 PM

P: n/a

I don't understand this conversion!
[tex]\sum_{n=1}^\infty \frac{sin(n\pi /2)}{n!} = \sum_{n=0}^\infty \frac{(1)^n}{(2n+1)!} [/tex] I know that the numerator of the left side is 0 when n is an even number. When n is odd, the numerator is either +1 or 1. But how do i continue? 


#2
Mar1106, 01:05 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

[tex]sin(\frac{n\pi}{2})= sin(\frac{(2m+1)\pi}{2})= (1)^m[/tex] so we have [tex]\sum_{n=1}^\infty \frac{sin(\frac{n\pi}{2})}{n!}= \sum_{m=0}^\infty\frac{(1)^m}{(2m+1)!}[/tex] Since m is a "dummy variable" (it just denotes a place in the series and doesn't appear in the final sum) just replace m by n to get the result you have the two "n"s on either side of the equation have different meanings. 


#3
Mar1106, 01:34 PM

P: n/a

How come the lower limit was replaced by 0 from 1?
Thanks. 


#4
Mar1206, 12:54 AM

P: 347

Series with Factorial
Because 2n+1 skips 1. So you could write 2n1 instead.



#5
Mar1206, 02:18 AM

HW Helper
P: 1,422

[tex]\sum_{n = 1} ^ {\infty} \left( \frac{\sin \left( \frac{n \pi}{2} \right)}{n!} \right) = \sum_{n = 0} ^ {\infty} \left( \frac{(1) ^ n}{(2n + 1)!} \right)[/tex] It's not only that n = 1 has been replaced by n = 0, but the n! in the denominator has also been replaced by (2n + 1)!. Do you notice this? As HallsofIvy has already pointed out: If n is even then [tex]\frac{\sin \left( \frac{n \pi}{2} \right)}{n!} = 0[/tex], right? So you'll be left with the terms with odd n only, now let n = 2m + 1 This means n is odd right? And since n >= 1 (the series starts from n = 1, and 1 is an odd number), so 2m + 1 >= 1, so m >= 0, which means the new series will start from m = 0. So change n to m, we have: [tex]\sum_{n = 1} ^ {\infty} \left( \frac{\sin \left( \frac{n \pi}{2} \right)}{n!} \right) = \sum_{m = 0} ^ {\infty} \left( \frac{\sin \left( \frac{(2m + 1) \pi}{2} \right)}{(2m + 1)!} \right) = \sum_{m = 0} ^ {\infty} \left( \frac{(1) ^ m}{(2m + 1)!} \right)[/tex] Can you get it now? :) 


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