Which number is greater? $2\sqrt{2}$ or $e$ I have to determine which number is greater, $2\sqrt{2}$ or $e$.
I had a similar question as well, it was $2^\sqrt{2}$ compared to $e$.
For that one I managed to prove the inequality by using the increasing sequence converging to $e$: $(1+\frac1n )^n $
So I just searched for a value to assign to n such that $(1+\frac1n )^n \gt 2^\sqrt2$
I tried to proceed in a similar way with $2\sqrt{2} \gt \lt e$ , but it seems I can't get nowhere (I used the sequence decreasing and converging to $e$:  $(1+\frac1n )^{n+1}$ )
Is there another way to prove the inequality without the use of the calculator and maybe using derivatives? The question was in a derivatives file, so I'm wondering is there's a way to get to the end using them.
Any hint would be much appreciated, thanks.
 A: 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 $4\le n+1$:
$$4^n\lt16n!\implies4^{n+1}=4\cdot4^n\lt4\cdot16n!\le(n+1)16n!=16(n+1)!$$
A: 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 $121\cdot8=968>900$.
A: $$\sum_{i=k}^\infty \frac {1}{i!} = \frac{1}{k!} \left(1+\frac{1}{k+1} + \frac{1}{(k+1)(k+2)} + \cdots \right)
\\ < \frac{1}{k!}\left( 1 + \frac{1}{k+1} + \frac{1}{(k+1)^2} + \cdots \right)
\\ = \frac{1}{k!} \frac{1}{1-\frac{1}{k+1}}=\frac{k+1}{k \cdot k!}$$
Therefore $$e = 2+ \sum_{i=2}^\infty \frac{1}{i!} < 2 + \frac{3}{2\cdot 2!} = \frac{11}{4} = \sqrt{\frac{121}{16}} < \sqrt{\frac{128}{16}}=2\sqrt{2}.\blacksquare $$
A: \begin{gather*}
3^5 = 243 < 256 = 2^8, \
\therefore\ \log_32 > \frac58; \\
\log_e3 = \log_e\left(1 + \frac12\right) - \log_e\left(1 - \frac12\right) > 2\left(\frac12 + \frac13\left(\frac12\right)^3\right) = \frac{13}{12}; \\
\therefore\ \log_e2 = \log_32 \cdot \log_e3 > \frac58\cdot\frac{13}{12} = \frac{65}{96} > \frac{64}{96} = \frac23, \
\therefore\ 2 > e^{2/3}, \
\therefore\ 2^{3/2} > e.
\end{gather*}
A: $$\ln (2×\sqrt 2)=\ln 2+\dfrac 12 ×\ln 2=\dfrac 32 \ln2$$
$$\ln 2 = \frac{2}{3} +\frac{3}{4}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(n+2)(n+3)}>\dfrac 23$$
$$ \implies \ln (2×\sqrt 2)>1=\ln e \Longrightarrow 2×\sqrt 2 >e.$$
A: You aleady received good and simple answers to the question.
Now just for your curiosity: during your steps, your tried to find $n$ such that
$$\left(1+\frac{1}{n}\right)^n > k$$
The only explicit solution for the equality is given in terms of Lambert funnction and it is
$$n=\frac{\log (k)}{W_{-1}\left(-\frac{\log (k)}{k}\right)+\log (k)}$$ For $k=2^\sqrt2$, the exact solution would be, as a real, $n=24.6624$, then $\lceil n \rceil=25$.
Computing
$$\left(\frac{26}{25}\right)^{25}=\frac{236773830007967588876795164938469376}{88817841970012523233890533447265625}\sim 2.66584$$ while $2^\sqrt2\sim 2.66514$.
