Which number is greater? $2\sqrt{2}$ or $e$

It is easy to show (by induction) that

$${2^n\over n!}\lt{1\over2^{n-4}}$$

for all $n\ge0$. It follows that

$$\begin{align} e^2&=1+2+{2^2\over2!}+{2^3\over3!}+{2^4\over4!}+{2^5\over5!}+\cdots\\ &=1+2+2+{4\over3}+{2\over3}+\left({2^5\over5!}+{2^6\over6!}+\cdots \right)\\ &=7+\left({2^5\over5!}+{2^6\over6!}+\cdots \right)\\ &\lt7+\left({1\over2}+{1\over4}+{1\over8}+\cdots\right)\\ &=7+1\\ &=8 \end{align}$$

so $e\lt\sqrt8=2\sqrt2$.

Remark: The induction proof for $2^n/n!\lt1/2^{n-4}$, best rewritten as $4^n\lt16n!$, requires verifying the first few "base" cases; the induction itself kicks in when $4\le n+1$:

$$4^n\lt16n!\implies4^{n+1}=4\cdot4^n\lt4\cdot16n!\le(n+1)16n!=16(n+1)!$$

Consider the series for $e^{-1}$, which is alternating. Then $$ e^{-1}>1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}=\frac{11}{30} $$ and $$ \frac{11}{30}>\frac{1}{2\sqrt{2}} $$ because $121\cdot8=968>900$.

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

Therefore $$e = 2+ \sum_{i=2}^\infty \frac{1}{i!} < 2 + \frac{3}{2\cdot 2!} = \frac{11}{4} = \sqrt{\frac{121}{16}} < \sqrt{\frac{128}{16}}=2\sqrt{2}.\blacksquare $$