Calculate $\sum_{n=1}^\infty\frac{1}{(n-1)!+n!}$

Question

$$\text{Calculate } \sum_{n=1}^\infty\frac{1}{(n-1)!+n!}$$

I'm having a hard time solving this. I now I must create a minus to transform it into a telescopic series but I've been staring at it for a long time and do not see where the minus is supposed to appear. I suppose I probably need to group $n!$ and $(n-1)!$ as $((n-1)!)(1+n)$ but, again, I still do not know what to do with the numerator. Any help is much appreciated!

Answer 1

Hint:

For $n(n+1)\ne0,$

$$\dfrac1{(n-1)!+n!}=\dfrac1{(n-1)!(1+n)}=\dfrac n{(n+1)!}=\dfrac{n+1-1}{(n+1)!}=\dfrac1{n!}-\dfrac1{(n+1)!}$$