# Rudin's explanation of how rapid the series 1/(n!) converges

1. May 6, 2013

In Rudin's Principle's of Mathematical Analysis, Rudin days that we can estimate how fast the series $\sum\frac{1}{n!}$ converges by the following:
Put
$$s_{n}=\sum_{k=0}^{n}\frac{1}{k!}.$$
Then
$$e-s_{n} =\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\cdots<\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots\right)=\frac{1}{n!n}$$
so that
$$0<e-s_{n}<\frac{1}{n!n}.$$
The part that bothers me is
$$\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots\right)=\frac{1}{n!n}.$$
Using Maple I was able to see that
$$\frac{1}{(n+1)!}\sum_{k=0}^{\infty}\frac{1}{(n+1)^{k}}=\frac{1}{n!n}$$
but what if I did not have access to anything like Maple or Mathematica. How would I be able to figure out that the equality holds?

2. May 6, 2013

### Office_Shredder

Staff Emeritus
That sum is a geometric series - do you know what to do with those?

3. May 6, 2013

Oh. HAHAHAHA. Wow. Okay. I see it now. Thanks.

4. May 6, 2013

### mathman

The sum is simply a geometric series with r = 1/(n+1), so the sum = 1/(1-r) = (n+1)/n.