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

How would I prove the following statement?

$\alpha _{n}< 3$

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    $\begingroup$ Hint: $\alpha_n<e=\sum_{k=0}^{\infty}\dfrac{1}{k!}<3$ $\endgroup$ – Kevin Song Jan 21 at 6:22
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$2! \geq 2, 3! > 2^{2}, ...$ Can you finish?

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

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