# How to prove this integer inequality? [duplicate]

Given $$\alpha _{n}= 1+\frac{1}{1!}+...+\frac{1}{n!} ,\forall n\in \mathbb{N}$$.

How would I prove the following statement?

$$\alpha _{n}< 3$$

• Hint: $\alpha_n<e=\sum_{k=0}^{\infty}\dfrac{1}{k!}<3$ – Kevin Song Jan 21 at 6:22

$$2! \geq 2, 3! > 2^{2}, ...$$ Can you finish?

[Use the fact that $$\sum\limits_{k=1}^{\infty} \frac 1 {2^{k}}=1$$].