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

Question

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.

Answer 1

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 $4\le n+1$:

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

Answer 2

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 $121\cdot8=968>900$.

Answer 3

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

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

Answer 4

\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*}

Answer 5

$$\ln (2×\sqrt 2)=\ln 2+\dfrac 12 ×\ln 2=\dfrac 32 \ln2$$

$$\ln 2 = \frac{2}{3} +\frac{3}{4}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)(n+2)(n+3)}>\dfrac 23$$

$$ \implies \ln (2×\sqrt 2)>1=\ln e \Longrightarrow 2×\sqrt 2 >e.$$

Answer 6

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 $2^\sqrt2\sim 2.66514$.