Series and factorial

[thechunk]

I’ve been playing around with the infinite series:
\sum_{k=1}^\infty \frac{k}{(k+1)!}

I haven’t really gotten anywhere with it however I punched it into my calculator and it determined the sum to be 1. And the sum of n terms of the series equals
1-\frac{1}{(n+1)(n!)}
Why is this so? Any help is much appreciated.

[joeboo]

Use induction on that last statement. Show its true for n = 1, then assume it's true for n = k, and show it's true for n = k+1

[thechunk]

I see how I can use induction to find why 1-\frac{1}{(n+1)(n!)}
gives the sum of the series but how would you analytically come up with that expression in the first place. My calculator did it in a second, how did it generate the expression. Is there something I am missing?

[shmoe]

It's a telescoping series, this may help:

\sum_{k=1}^{n}\frac{k}{(k+1)!}=\sum_{k=1}^{n}\left (\frac{k+1}{(k+1)!}-\frac{1}{(k+1)!}\right)


For the infinite series you can also consider:

\frac{d}{dx}\left(\frac{e^x-1}{x}\right)=\sum_{k=1}^{\infty}\frac{kx^{k-1}}{(k+1)!}

[thechunk]

Thanks shmoe, I lost my negative and made the series, dare I say, even more infinite. Mwahahaha...