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.  
 A: $$\log(2)=\int_{0}^{1}\frac{dx}{1+x}\stackrel{\text{Holder}}{<}\sqrt[3]{\int_{0}^{1}\frac{dx}{(1+x)^{9/8}}\int_{0}^{1}\frac{dx}{(1+x)}\int_{0}^{1}\frac{dx}{(1+x)^{7/8}}} $$
leads to a stronger inequality than $\log(2)<3^{-1/3}$.
A: 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.
A: Hint:
Rewrite $\ln 2\;$ as $\;\ln\biggl(\dfrac{1+\frac13}{1-\frac13}\biggr)$, and  note
$$\ln\biggl(\frac{1+x}{1-x}\biggr)=2\biggl(x+\frac{x^3}3+\frac{x^5}5+\dotsm\biggr)\quad\text{for }\;\lvert x\rvert<1.$$
A: 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 ...
A: 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 .
A: with $0<\alpha<0.0008$ we have
$$\ln2=\int_1^2\frac{1}{x}dx\leq\int_1^2\frac{1}{x^{1-\alpha}}dx=\frac{2^\alpha-1}{\alpha}<\frac{1}{\sqrt[3]{3}}$$
A: $\ln2<{1\over{3^{1/3}}}$
$e^{\ln2}=2<e^{{1\over{3^{1/3}}}}$
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 + $\sum_{n=N}^\infty$ $\left({1\over{3^{1/3}}}\right)^n{1\over{n!}}$
