How to compare the values $\sqrt 2$ and $\ln(3).$ How to compare the values $\sqrt 2$ and $\ln(3)?$ I know only $\ln(x)<x$ and $\sqrt{x}<x$. Please help. Thanks.

Here's another series argument:
Use the fact that $\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} -\cdots$ and $\ln(2+x) = \ln 2 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{24} \cdots$. The first expansion is obtained by successfully differentiating $(1+x)^n$ and substituting $0$ with $n = \frac{1}{2}$. 
Then $\sqrt{2} - \ln 3$ equals:
$$\left(1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} \cdots \right) - \left(\ln2 
 + \frac{1}{2} - \frac{1}{8} + \frac{1}{24} \right)$$
$$=\frac{49}{48} - \ln 2$$
and approximating $\ln 2$ as $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12}$, we have:
$$\frac{49}{48} - \frac{7}{12} = \frac{7}{16} $$
The Lagrange error bound for $\sqrt{1+x}$ is:
$$|R_n| ≤ \frac{|1/16|(1-0)^4}{4!} \Rightarrow |R_n| ≤ \frac{1}{384} (\approx 0.0026)$$
and the Lagrange error bound for $\ln(1+x)$ is:
$$|R_n| ≤ \frac{|1/24|(1-0)^4}{4!} \Rightarrow |R_n| ≤ \frac{1}{576} (\approx 0.0017)$$
Meanwhile, the maximum error for approximating $\ln 2$ to order $x^4$ is:
$$|R_n| ≤ \frac{|-1/4|(1-0)^5}{5!} \Rightarrow |R_n| ≤ \frac{1}{480} (\approx 0.0021)$$
In the worst possible scenario, adding up all these errors gets nowhere near $\frac{7}{16}$. Therefore $\sqrt{2} - \ln 3 > 0$ and $\sqrt{2} > \ln 3$.
