Prove that $\ln2<\frac{1}{\sqrt[3]3}$

Question

Prove that $$\ln2<\frac{1}{\sqrt[3]3}$$ without calculator.

Even $\ln(1+x)\leq x-\frac{x^2}{2}+\frac{x^3}{3}-...+\frac{x^{51}}{51}$ does not help here and we need another Taylor.

Answer 1

$$\log(2)=\int_{0}^{1}\frac{dx}{1+x}\stackrel{\text{Holder}}{<}\sqrt[3]{\int_{0}^{1}\frac{dx}{(1+x)^{9/8}}\int_{0}^{1}\frac{dx}{(1+x)}\int_{0}^{1}\frac{dx}{(1+x)^{7/8}}} $$ leads to a stronger inequality than $\log(2)<3^{-1/3}$.

Answer 2

Setting $s=\sqrt[3]{3}$, you can try seeing whether $2<e^{1/s}$ by using a suitable truncation of the Taylor series. At degree $5$ we have $$ 2<1+\frac{1}{s}+\frac{1}{2s^2}+\frac{1}{18}+\frac{1}{72s}+\frac{1}{360s^2} $$ that is, $$ 360s^2<360s+180+20s^2+5s+1 $$ or $$ 340s^2-365s-181<0 $$ which is satisfied so long as $$ s<\frac{365+\sqrt{379385}}{680} $$ Now proving that $$ \left(\frac{365+\sqrt{379385}}{680}\right)^3>3 $$ is just (very) tedious computations, but they don't need more than pencil and paper.

Answer 3

Inequalities like this can obviously be "proved" by plugging numbers into a scientific calculator, which means they can also be established, at least in principle, invoking pretty much any convergent Taylor series for the functions involved, with appropriate error bounds. The challenge is organize things so that the arithmetic stays manageable. Here is one attempt to do so.

It's convenient to begin by noting that

$$\ln2\lt{1\over\sqrt[3]3}\iff3\ln2\lt\sqrt[3]9=2\left(1+{1\over8}\right)^{1/3}$$

To get started, we have

$$\begin{align} \ln\left(1+x\over1-x\right)&=\ln(1+x)-\ln(1-x)\\ &=\left(x-{1\over2}x^2+{1\over3}x^2-\cdots\right)+\left(x+{1\over2}x^2+{1\over3}x^3+\cdots\right)\\ &=2\left(x+{1\over3}x^3+{1\over5}x^5+{1\over7}x^7+\cdots\right)\\ &\le2x+{2\over3}x^3+{2\over5}x^5+{x^7\over3(1-x)} \end{align}$$

(where we've generously changed the $7$ to a $6$ and bounded the remainder with a geometric series). Thus

$$3\ln2=3\ln\left(1+{1\over3}\over1-{1\over3} \right)\le2+{2\over3^3}+{2\over5\cdot3^4}+{1\over2\cdot3^6}=2+{20\cdot3^3+4\cdot3^2+5\over2\cdot5\cdot3^6}\\\lt2+{20\cdot3^3+4\cdot3^2+6\over2\cdot5\cdot3^6}=2+{90+6+1\over5\cdot3^5}=2+{97\over3^2\cdot135}$$

On the other hand

$$(1+x)^{1/3}=1+{1\over3}x-{1\over9}x^2+{5\over81}x^3-\cdots\ge1+{1\over3}x-{1\over9}x^2$$

and thus

$$2\left(1+{1\over8}\right)^{1/3}\ge2+{1\over3\cdot4}-{1\over3^2\cdot32}=2+{3\cdot8-1\over3^2\cdot32}=2+{23\over3^2\cdot32}$$

It follows that $\ln2\lt1/\sqrt[3]3$ if $97/135\lt23/32$. This can be finished off with some straightforward multiplication. But it's easier (or more fun) to check that

$${97\over135}\lt{23\over32}\iff{38\over97}\gt{9\over23}\iff{21\over38}\lt{5\over9}\iff{17\over21}\gt{4\over5}\iff85\gt84$$

Answer 4

Hint:

Rewrite $\ln 2\;$ as $\;\ln\biggl(\dfrac{1+\frac13}{1-\frac13}\biggr)$, and note $$\ln\biggl(\frac{1+x}{1-x}\biggr)=2\biggl(x+\frac{x^3}3+\frac{x^5}5+\dotsm\biggr)\quad\text{for }\;\lvert x\rvert<1.$$

Answer 5

Using $-\ln(1-x)=x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\ldots$ (cf. Adren's comment) we have $$\ln 2=-\ln\frac12=\sum_{n=1}^\infty\frac{1}{n2^n}.$$ We can estimate the tail $$\sum_{n=N}^\infty\frac{1}{n2^n}<\sum_{n=N}^\infty\frac{1}{N2^n}=\frac1{N2^{N-1}} $$ "For no apparent reason", we pick $N=10$ and see $$\ln 2<\sum_{n=1}^9\frac1{n2^n}+\frac1{10\cdot 2^9} =\frac{447173}{645120}.$$ Raising the right hand side to the third power proves the desired result: $$\frac{447173}{645120}=\frac{89418364010966717}{268485921865728000}=\frac13-\frac{76943277609283}{268485921865728000}. $$ Now if only I could convince you that I did all the calculations by hand ...

Answer 6

Let $x\geq1$ then. $$\ln{x}\leq(x-1)\sqrt[3]{\frac{2}{x^2+x}}$$

Apparently found by user Michael Rozenberg here Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$

The application is direct .

Answer 7

with $0<\alpha<0.0008$ we have $$\ln2=\int_1^2\frac{1}{x}dx\leq\int_1^2\frac{1}{x^{1-\alpha}}dx=\frac{2^\alpha-1}{\alpha}<\frac{1}{\sqrt[3]{3}}$$

Answer 8

$\ln2<{1\over{3^{1/3}}}$

$e^{\ln2}=2<e^{{1\over{3^{1/3}}}}$

Using the definition of exponential function (by sum of infinit sequence) on the right side:

$e^{{1\over{3^{1/3}}}}=\sum_{n=0}^\infty$ $\left({1\over{3^{1/3}}}\right)^n{1\over{n!}}$

Let us devide the sum into two parts and decrease it by using N (fix positiv integer number) instead of first n places of the n!.

$\sum_{n=0}^\infty$ $\left({1\over{3^{1/3}}}\right)^n{1\over{n!}}$=1+$\sum_{n=1}^\infty$ $\left({1\over{3^{1/3}}}\right)^n{1\over{n!}}$>$1+\sum_{n=1}^N$ $\left({1\over{N\cdot3^{1/3}}}\right)^n$+$\sum_{n=N}^\infty$ $\left({1\over{3^{1/3}}}\right)^n{1\over{n!}}$

Let us calculate the sum of the first N items (geometric sequence).

We can see that $S_N=$${\left({1\over{N\cdot3^{1/3}}}\right)^N-1}\over{\left({1\over{N\cdot3^{1/3}}}\right)-1}$ >1 for all N>2 and the limit of it goes to 1 if the N goes to infinity.

So: $1+\sum_{n=1}^N$ $\left({1\over{N\cdot3^{1/3}}}\right)^n$=1+$lim \over N- \infty$ ${\left({1\over{N\cdot3^{1/3}}}\right)^N-1}\over{\left({1\over{N\cdot3^{1/3}}}\right)-1}$=2

We obtaine that the right side of the statement lager than 2.

2 < 2 + $\sum_{n=N}^\infty$ $\left({1\over{3^{1/3}}}\right)^n{1\over{n!}}$