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

[DeadOriginal]

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?

 

[Office_Shredder]

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

 

[DeadOriginal]

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

 

[mathman]

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?

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