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

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.

• $\ln$ converges slowly. But even sowing $2<\exp(\frac1{\sqrt[3]3}$ requires at lest 6th order Feb 11, 2017 at 16:56
• @HagenvonEitzen I get it at order $5$, but it's tedious nonetheless. Feb 11, 2017 at 16:57
• You could maybe use $\ln(2)=\sum_{n=1}^\infty\frac{1}{n2^n}$ which converges quite rapidly ... and $\frac1{\sqrt[3]{3}}=\sum_{n=0}^\infty\binom{1/3}{n}\left(-\frac23\right)^n$ The former is obtained via Taylor's integral formula, so we can hope for an estimate of the queue of the series, expressed as an integral. The latter is an alternate series, and we know an estimation of its queue too. Feb 11, 2017 at 17:03
• @Adren it gives $\ln2>...$ and we need $\ln2<...$. Feb 11, 2017 at 17:10
• @MichaelRozenberg: Not sure ... We have, for all $n\ge1$ : $\ln(2)=\sum_{k=1}^n\frac{1}{k2^k}+\int_0^{1/2}\frac{(\frac12-t)^n}{(1-t)^{n+1}}\,dt$ And it should be possible to get an upper bound for the last integral. Feb 11, 2017 at 17:23

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

• I have no idea how you pull these out of thin air, +1
– user223391
Feb 12, 2017 at 1:25
• Maybe I am missing something, but cancelling the $\log(2)$ on the inside of the cube root with the $\log(2)$ on the left, then raising both sides to the $\frac32$ power, gives $\log(2)\le8\left(2^{1/16}-2^{-1/16}\right)$, but the right side is greater than $3^{-1/3}$. If the exponents were $10/9$ and $8/9$, then we would have $\log(2)\le9\left(2^{1/18}-2^{-1/18}\right)\lt3^{-1/3}$, but how to show the right-hand inequality?
– robjohn
Sep 23, 2021 at 20:39

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.

• Several pages of pencil and paper..Are we using Newton's Method to approximate the root? Feb 11, 2017 at 17:12
• @S.C.B. This is just integer numbers. Feb 11, 2017 at 17:19
• @S.C.B. Note that $\left(\frac{365+\sqrt N}{680}\right)^3>3$ iff $3\cdot 680^3<(365+\sqrt N)^3=365^3+3\cdot 365^2\sqrt N+3\cdot 365\cdot N+N\sqrt N$, i.e., to $779060\sqrt N> 47924300$, or $606934483600\cdot 379385 > 606934483600^2$. That's doable - on a long winter evening Feb 11, 2017 at 17:19
• @HagenvonEitzen OK, that's actually probably more feasible. I didn't think of it. It does seem my attention and cognitive ability generally decrease as the time I am awake increases. Feb 11, 2017 at 17:21
• @S.C.B. Though I'm sure that cognitive abilities of anybody would severely decrease when attempting to actually do those computations by hand :) Feb 11, 2017 at 17:34

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

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

Hint:

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

• This can be used to show $\ln2$ is greater than something, not less than. Feb 11, 2017 at 17:35
• @Bernard What estimate for $1/\sqrt[3]{3}$ are you intending to use that is less than $1/\sqrt[3]{3}$ and greater than an upper bound for $\ln(2)$?
– user366533
Feb 11, 2017 at 17:37
• @egreg Well, the tail can be estimated $\sum_{n=N}^\infty \frac{x^{2n+1}}{2n+1}<\sum_{n=N}^\infty \frac{x^{2n+1}}{2N+1}=\frac1{2N+1}\cdot\frac{x^{2N+1}}{1-x^2}$. Hence the error when stopping at $x^5$ as bernard did, is $<\frac17\cdot\frac{(1/3)^7}{1-1/9}=\frac1{13608}$ turns out as just good enough Feb 11, 2017 at 17:40
• @Hagen von Eitzen Are you saying to cube the series for $\ln(\frac{1+x}{1-x})$, truncate it after the seventh power of $x$, and substitute $1/3$ for $x$? Compare the number obtained from this computation with $1/3$?
– user366533
Feb 11, 2017 at 18:08

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 ...

• Looks good ! you forgot to write the exponent 3 (on last line) Feb 11, 2017 at 17:39

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 .

• Thank you! Now I know it. Sep 11, 2021 at 18:09
• @MichaelRozenberg Well the main idea is yours .Anyway your are welcome :-)! Sep 11, 2021 at 18:16

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

• How do you compute $2^\alpha$ without a calculator? Feb 11, 2017 at 17:52

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

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

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