Showing that $e^{-2} < \ln 2$

Question

I have to prove the following inequality: $e^{-2} < \ln2.$

Using Bernoulli's inequality, I showed that $2 \leq e$, and using this result I tried to simplify the inequality by using an upper estimate for $e^{-2}$ and a lower estimate for $\ln2$. Using this method, I got that $e^{-2} \leq \frac{1}{4}$, but I couldn't proceed from that point.

Any hints would be much appreciated.

Answer 1

Assuming you are allowed to use that $e>2$, you have that $$e^2>2^2=4$$ and therefore$$2^{(e^2)}>2^4=16$$ thus $$\ln 2^{(e^2)} > \ln 16 > \ln e =1$$ (however, here I used that $\ln$ is a monotone increasing function). Now since the left hand side of the above inequality can be written as $e^2\cdot \ln 2$ you have that $$e^2 \ln 2> 1$$ which gives $$\ln 2 > \frac{1}{e^2}=e^{-2}$$

Answer 2

If you're careful enough to handle the rounding errors properly on your domain, you could try using some inequality work with some Taylor series.

For $x<0$, $\space e^x < 1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} $

$(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4<\log(x)$

Now plug in $x=-2$ and $x=2$ in their respective places:

$e^{-2}< 1-2+2-\frac{8}{6}+\frac{16}{24} =\frac{1}{3} < \frac{7}{12} =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}<\log(2)$

Thus, it checks out that $e^{-2}<\log(2)$.

Answer 3

It seems I posted my question too early, since I found the answer after 2 minutes. Here it is: According to the previous results, we only need to show that $$ \frac{1}{4} \leq \ln2. $$ But $$\frac{1}{4} = \ln e^{\frac{1}{4}}$$ therefore - using the strict monotinicity of the $\ln$ function - we only need to prove that $$e^{\frac{1}{4}} \leq 2 \Longleftrightarrow \sqrt[4]{e} \leq \sqrt[4]{2^4}.$$ Using the strict monotonicity of the $\sqrt[4]{}$ function the inequality simplifies to $$ e \leq 16. $$ But $$ e = \sum_{n = 0}^\infty \frac{1}{n!} < 1 + \sum_{n = 0}^\infty \left(\frac{1}{2}\right)^n = 3 < 16. $$

Answer 4

This is an overkill, but I think it is interesting, anyway. We have that $f(x)=e^x \log(1+x)$ is a convex increasing function over $[0,1]$, since $f'(x)=\frac{e^x}{1+x}+f(x)$. Jensen's inequality hence gives: $$ \forall x\in(0,1],\qquad f(x)> f(0)+f'(0)\, x = x, $$ so $f(1)>1$ gives $e\log 2>1$ and $e^{-1}<\log 2$.

The argument can be improved, too. By showing that $f(x)>x+\frac{x^2}{2}+\frac{x^3}{3}$ it follows that $e\log 2>\frac{11}{6}$, so $e^{-1}<\frac{6}{11}\log 2$ and $e^{-2}<\frac{36}{121}\log^2 2<\frac{3}{10}\log^2 2.$

Answer 5

Yet another argument, using the inequalities $2<e<4$: From $e<4$ we get $1<2\ln2$ so $\ln2>1/2$; from $2<e$ we get $1/2>1/e$. Conclusion: $\ln2>e^{-1}$.