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.
 A: If $n \geq 0$, you have
$$ \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \frac{1}{(n+2)(n+3)(n+4)} + \cdots$$
$$ \leq
  \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \cdots $$
$$ < \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots = 1 < 2$$
A: Hint: Compare the left side to a geometric series.
A: 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 $n+1$
A: Well... you've finished!!!
Assume it's true...
$\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots < \frac{3}{(n+1)!}$
Now multiplying both side for $(n+1)!$  follows that
$1 + \frac{1}{(n+2)} + \frac{1}{(n+2)(n+3)} + \cdots... < 3$ (1)
Observing that for fixed positive $n$ the inequality $\frac{1}{n + 2 + k} < \frac{1}{2}$ for each $k$ positive the left side of (1) is of course less than
$1 + \frac{1}{(n+2)} + \frac{1}{(n+2)(n+3)} + \cdots... < \sum_{k=0}^{+\infty} \left( \frac{1}{2} \right)^k = 2 < 3$
Then our assumption was correct!!!
