Help with difficult telescoping series question: $\frac3{1!+2!+3!}+\frac4{2!+3!+4!}+\ldots+\frac{2012}{2010!+2011!+2012!}$

Question

Evaluate

$$\frac3{1!+2!+3!}+\frac4{2!+3!+4!}+\ldots+\frac{2012}{2010!+2011!+2012!}\;.$$

I see that the question is telescoping, but I don't know how to break it down into a form similar to that of the most basic telescoping series. What would be the best method to simplify this question?

Answer 1

The denominator of each term is $$(n-2)!+(n-1)!+n!=(n-2)!(1+n-1+(n-1)n) = (n-2)!\,n^2,$$ so each term simplifies to $$\frac{n}{(n-2)!n^2}=\frac{1}{(n-2)!n}=\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!},$$ and now you can see that the series telescopes.