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?

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 $\ln(3/2) < \dfrac{13}{32}.$
