# Series and factorial

1. Mar 29, 2005

### 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.

2. Mar 29, 2005

### 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

3. Mar 29, 2005

### 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?

4. Mar 29, 2005

### 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:

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

5. Mar 29, 2005

### thechunk

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

6. Oct 31, 2010

### Lionel 1987

Whay about : \sum_{n=1}^{\infty}\frac{8^{n}}{(n)!} ( I copy like this cause i don´t know how to put the symbol)Does anybody know how to solve this? PLease, help.