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!
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.
$$\frac{1}{3}+\frac{1}{8}+\frac{1}{30}+\frac{1}{144}+\frac{1}{840}+\cdots=\frac{1}{2}$$

$$\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+\frac{1}{30}+\frac{1}{144}+\cdots=\frac{1}{2}+\frac{1}{2}=1$$