Find the sum of the series of $\frac{ 1}{n(n-2)!}$ from $n=2$ to infinity I am trying to find the sum of the series from $n=2$ to infinity of $\frac{ 1}{n(n-2)!}$ 
I was originally given: 1/2 + 1/3 + 1/8 + 1/30 + 1/144 ... ? and I had to figure out the next term (which is 1/840) and then figure out the sum of the series. I deduced what the formula is to make the series but I have no clue how to find the sum. (Brute forcing the sum on a calculator gave me a number very very close to 1 so I'm guessing that's the sum, but I don't know how to find that sum using calculus) Please help!
 A: This is
$$\sum_{n=2}^\infty\frac{(n-1)}{n!}
=\sum_{n=2}^\infty\frac{n}{n!}-\sum_{n=2}^\infty\frac{1}{n!}
=\sum_{n=2}^\infty\frac{1}{(n-1)!}-\sum_{n=2}^\infty\frac{1}{n!}
=\sum_{m=1}^\infty\frac{1}{m!}-\sum_{n=2}^\infty\frac{1}{n!}
$$
etc.
A: You already accepted the answer, which is fine. Here is just another way of looking at it:
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....=\sum_{n=0}^\infty\frac{x^n}{n!}
$$
Now when you divide everything by $x$ and then take derivative on both sides, you end up with:
$$\frac{xe^x-e^x}{x^2}=\frac{-1}{x^2}+1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}....=\sum_{n=0}^\infty\frac{(n-1)x^{n-2}}{n!}$$
When you plug in $x=1$ and move over $\frac{-1}{x^2}$ to the left side, you arrive at your desired sum, which is indeed $1$ as you suggested.
A: $$\dfrac{n-1}{n!}=f(n-1)-f(n)$$ where $f(x)=\dfrac1{(x-1)!}$
$$\sum_{n=2}^m\dfrac{n-1}{n!}=f(1)-f(m)$$
Now $\lim_{m\to\infty}f(m)=?$
See also: How to solve this summation without taylor?
A: $$\frac{1}{3}+\frac{1}{8}+\frac{1}{30}+\frac{1}{144}+\frac{1}{840}+\cdots=\frac{1}{2}$$
$$\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+\frac{1}{30}+\frac{1}{144}+\cdots=\frac{1}{2}+\frac{1}{2}=1$$
