Showing that $\sum_{k=N+1}^\infty \frac {1}{k!} < \frac {1}{N!}$

Question

I was constructing a proof through inequalities, but I am having a bit of problem showing the following step: $$\sum_{k=N+1}^\infty \frac {1}{k!} < \frac {1}{N!}$$

Is there any quick way to show this?

Answer 1

$$\sum_{k=N+1}^\infty \frac {1}{k!} =\frac{1}{N!}\left(\frac{1}{N+1}+\frac{1}{(N+1)(N+2)}+...\right)<$$ $$<\frac{1}{N!}\left(\frac{1}{N+1}+\frac{1}{(N+1)^2}+...\right)=\frac{1}{N!}\cdot\frac{\frac{1}{N+1}}{1-\frac{1}{N+1}}\leq \frac {1}{N!}$$

Answer 2

All we need to prove is :

$$\sum_{k=N+1}^\infty \frac {N!}{k!} < 1$$

Now, since

$$\frac{1}{N+1}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+1)(N+2)(N+3)} \ldots <\frac{1}{(N+1)}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+2)(N+3)}$$

Can you see the telescoping sum now, doing that we get

$$\frac{1}{N+1}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+1)(N+2)(N+3)} \ldots <\frac{2}{N+1} < 1 ~\forall ~N \ge 2$$

Answer 3

An alternative proof:

$$\sum_{k=N+1}^\infty\frac1{k!}<\sum_{k=N+1}^\infty\frac{k-1}{k!}=\sum_{k=N+1}^\infty\frac1{(k-1)!}-\sum_{k=N+1}^\infty\frac1{k!}=\frac1{N!}$$