How to prove this integer inequality?

Question

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.S Jan 21 '20 at 06:22 — Kevin.S, Jan 21 '20 at 06:22

score 1 · Answer 1 · answered Jan 21 '20 at 06:22

1

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

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

answered Jan 21 '20 at 06:22

Kavi Rama Murthy

1 Answers1