Prove inquality involving factorial This is part of an analysis problem I'm working on.
Show that:
$ \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + ... < \frac{3}{(n+1)!}$
After some algebra I got
$ \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \frac{1}{(n+2)(n+3)(n+4)} + ... < 2$
Now I'm stuck.

By induction:
$e^x=1+x+x^2/2+x^3/3!+...+x^n/n!+...$
So if $x=1$
$e=2+1/2+...+1/n!+...$
If $n=0$ (in your formula) we get:
$3>1+1/2+...+1/n!$
which is clearly true since $3>e-1$. 
If for $n$ is true then:
$\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + ... < \frac{3}{(n+1)!}$
so:
$\frac{1}{(n+2)!} + \frac{1}{(n+3)!} + ... <\frac{1}{(n+2)}\frac{1}{(n+1)!} +\frac{1}{(n+2)}\frac{1}{(n+2)!}+ ... < \frac{1}{(n+2)}\frac{3}{(n+1)!}=\frac{3}{(n+2)!}$
and we've proved the statement for $n+1$
