# Series with Factorial

by dekoi
Tags: factorial, series
 P: n/a I don't understand this conversion! $$\sum_{n=1}^\infty \frac{sin(n\pi /2)}{n!} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}$$ 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?
Math
Emeritus
Thanks
PF Gold
P: 39,345
 Quote by dekoi I don't understand this conversion! $$\sum_{n=1}^\infty \frac{sin(n\pi /2)}{n!} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}$$ 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?
AS you say, if n is even then sin(n$\pi$)= 0. Let n= 2m+1 so that for every m= 1, 2, 3,... n is odd. Notice that $sin(\frac{\pi}{2})= 1$, $sin(\frac{3\pi}{2})= -1$, $sin(\frac{5\pi}{2})= 1$ etc. In other words, $sin(\frac{(2m+1)\pi}{2}$ is 1 if m even (0, 2, etc.) and -1 is m is odd.
$$sin(\frac{n\pi}{2})= sin(\frac{(2m+1)\pi}{2})= (-1)^m$$
so we have
$$\sum_{n=1}^\infty \frac{sin(\frac{n\pi}{2})}{n!}= \sum_{m=0}^\infty\frac{(-1)^m}{(2m+1)!}$$

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.
 P: n/a How come the lower limit was replaced by 0 from 1? Thanks.
 P: 347 Series with Factorial Because 2n+1 skips 1. So you could write 2n-1 instead.
HW Helper
P: 1,421
 Quote by dekoi How come the lower limit was replaced by 0 from 1? Thanks.
Hmmm, look at the equality again:
$$\sum_{n = 1} ^ {\infty} \left( \frac{\sin \left( \frac{n \pi}{2} \right)}{n!} \right) = \sum_{n = 0} ^ {\infty} \left( \frac{(-1) ^ n}{(2n + 1)!} \right)$$
It's not only that n = 1 has been replaced by n = 0, but the n! in the denominator has also been replaced by (2n + 1)!.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
$$\frac{\sin \left( \frac{n \pi}{2} \right)}{n!} = 0$$, 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:
$$\sum_{n = 1} ^ {\infty} \left( \frac{\sin \left( \frac{n \pi}{2} \right)}{n!} \right) = \sum_{m = 0} ^ {\infty} \left( \frac{\sin \left( \frac{(2m + 1) \pi}{2} \right)}{(2m + 1)!} \right) = \sum_{m = 0} ^ {\infty} \left( \frac{(-1) ^ m}{(2m + 1)!} \right)$$
Can you get it now? :)

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