Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$
My try:
we have $$\frac{13}{32}=\frac{2^2+3^2}{2^5}=\frac{1}{8}\left(1+(1.5)^2)\right)$$
Let $x=1.5$
Now consider the function $$f(x)=\frac{1+x^2}{8}-\ln x$$
$$f'(x)=\frac{x}{4}-\frac{1}{x}$$ So $f$ is Decreasing in $(0,2)$
any help here?
 A: \begin{align*}
\exp\left(\frac{13}{64}\right) & = \exp\left(\frac15\right)\exp\left(\frac1{320}\right) \\
& > \left(1 + \frac15 + \frac1{50} + \frac1{750}\right)\left(1 + \frac1{320}\right) \\
& = \left(1 + \frac{166}{750}\right)\left(1 + \frac1{320}\right) \\
& = \frac{458}{375}\times\frac{321}{320} = \frac{229\times107}{125\times160} \\
& = \frac{24{,}503}{20{,}000} > \frac{24{,}500}{20{,}000} = \frac{49}{40} \\
\therefore\
\exp\left(\frac{13}{32}\right) & > \left(\frac{49}{40}\right)^2 = \frac{2{,}401}{1{,}600} > \frac32.
\end{align*}
A: The difference is so small that I see no other way than to do the computation.  Note $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$ implies $$e^{13/32} > 1 + \frac{13}{32} + \frac{(13/32)^2}{2!} + \frac{(13/32)^3}{3!} + \frac{(13/32)^4}{4!} = \frac{12591963}{8388608} > \frac{3}{2}.$$
A: We'll prove that $\ln\frac{3}{2}<\frac{13}{32},$ for which we'll prove that for any $x\geq1$ the following inequality holds.
$$\ln{x}\leq(x-1)\sqrt[3]{\frac{2}{x^2+x}}.$$
Indeed, let $f(x)=(x-1)\sqrt[3]{\frac{2}{x^2+x}}-\ln{x}.$ 
Thus,
$$f'(x)=\frac{\sqrt[3]2(x^2+4x+1)-3\sqrt[3]{x(x+1)^4}}{3\sqrt[[3]{(x^2+x)^4}}=\frac{2(x^2+4x+1)^3-27x(x+1)^4}{someting\\positive}=$$
$$=\frac{(2x^2+5x+2)(x-1)^4}{someting\\positive}\geq0,$$
which gives $$f(x)\geq f(1)=0.$$
Thus, $$\ln1.5<0.5\sqrt[3]{\frac{2}{3.75}}=\frac{1}{\sqrt[3]{15}}.$$
Id est, it's enough to prove that:
$$\frac{1}{\sqrt[3]{15}}<\frac{13}{32}$$ or
$$32768<32955$$ and we are done!
A: I want to point out that $\frac{13}{32}$ is the value of the Pade $(2,1)$ approximation of $\ln (1+x)$ at $x=\frac{1}{2}$.
In detail, the Pade $(2, 1)$ approximation of $\ln (1+x)$ is $g(x) = \frac{x^2+6x}{6+4x}$.
It is easy to prove that $\frac{x^2+6x}{6+4x} > \ln (1+x)$ for $x > 0$.
Indeed, let $f(x) = \frac{x^2+6x}{6+4x} - \ln(1+x)$. We have
$f'(x) = \frac{x^3}{(3+2x)^2(1+x)} > 0$ for $x > 0$. Also, $f(0) = 0$. The desired result follows.
We have $g(\frac{1}{2}) = \frac{13}{32} > \ln \frac{3}{2}$.
We may find the Pade approximation by hand. We may first try the Pade $(1, 1)$ approximation, second try the Pade $(2, 1)$ approximation,
third try the Pade $(1, 2)$ approximation, and so on, until we find enough approximation.
More details about the Pade $(2, 1)$ approximation:
The Taylor expansion of $\ln(1+x)$ is $x - \frac{1}{2}x^2 + \frac{1}{3} x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 + \cdots$
Let $g(x) = \frac{a_0 + a_1x+ a_2x^2}{1 + b_1x}$. Comparing the coefficients of $x^k$ for $k=0, 1, 2, 3$ of
$$(x - \frac{1}{2}x^2 + \frac{1}{3} x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 + \cdots)(1 + b_1x)
= a_0 + a_1x+ a_2x^2, $$
we obtain $a_0 = 0, a_1 = 1, a_2 = \frac{1}{6}, b_1 = \frac{2}{3}$. Then, $g(x) = \frac{x^2+6x}{6+4x}$.
A: I think we can try with $$\log\frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\dots\right)$$ and put $x=1/5$. The calculations are simple and one can estimate the error easily. Estimation of error gives $$\log\frac{1+x}{1-x}<2x+\frac{2x^3}{3}+\frac{2x^5}{5(1-x^2)}$$ With a little calculation of the easy variety (division by 2,3,5 etc) you can conclude that the right hand side of the above inequality for $x=1/5$ is less than $13/32$. 
A: You can use the series
$$\ln(3/2) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^nn} = \frac{1}{2} - \frac{1}{8} + \frac{1}{24} - \frac{1}{64} + \dotsc$$
Clearly the partial sums fluctuate closer and closer to $\ln(3/2)$, and each new term has a strictly smaller magnitude than the previous one. Thus, since
$$\sum_{n=1}^{6}\frac{(-1)^{n+1}}{2^nn} = \frac{259}{640} < \frac{909}{2240} = \sum_{n=1}^{7}\frac{(-1)^{n+1}}{2^nn} < \frac{13}{32},$$ it follows that $\ln(3/2) < \dfrac{13}{32}.$
