How do I prove that $\ln n > 1$ when $n > 2$? I've tried looking at the Taylor series of the function $\ln x$, which is $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (x-1)^n,$$ but wasn't able to extract any useful information out. I haven't tried induction but even if it's possible to prove this via induction, I'd rather see a constructive proof, if it's not too much to ask. Thanks for your help.
Edit: Here $n$ is an integer.
 A: Note that for the logarithm to base $b$, $\log_b b = 1$.
The logarithm (to any base) is a strictly monotonous function and so $\log_b a>1$ if $a>b$.
A: Hint: $\displaystyle\frac{d}{dx}\ln x=\frac1x>0$ for $x>2$, so the function is monotonically increasing. It suffices to show the conclusion for $n=3$.
A: If you are allowed to use the fact that $e < 3$ you can reason as follows:


*

*$\ln x$ is the inverse function of the strictly increasing function $e^x$. So, it is also strictly increasing.

*$\ln 3 > 1 \Leftrightarrow 3 > e$, which is true.

*Per induction you have $\ln (n+1) = \ln (n(1+\frac{1}{n})) = \ln n + \underbrace{\ln ((1+\frac{1}{n}))}_{>0}$
A: Here's one way to do it using the OP's Taylor expansion (and the fact that $\ln x$ is an increasing function of $x$):
If $n\ge3$. then
$$\ln n\ge\ln3\gt\ln2.89=2\ln1.7=2\left(0.7-{(0.7)^2\over2}+{(0.7)^3\over3}-{(0.7)^4\over4}+\cdots \right)\gt2\left(0.7-{(0.7)^2\over2}+{(0.7)^3\over3}-{(0.7)^4\over4}\right)$$
It remains to calculate
$$=0.7-{(0.7)^2\over2}+{(0.7)^3\over3}-{(0.7)^4\over4}\approx0.5093\gt{1\over2}$$
Remark: If you're willing to check that $3\gt(1+1/\sqrt2)^2$, then you can avoid the final messy calculation, using instead that
$$\ln3\gt2\left({1\over\sqrt2}-{1\over4}+{1\over6\sqrt2}-{1\over16} \right)={28\sqrt2-15\over24}\gt{28\cdot1.4-15\over24}={39.2-15\over24}\gt1$$
Added later: Even easier is
$$\ln3\gt\ln2.89=2\ln1.7\gt2\ln1.69=4\ln1.3\gt4\left(0.3-{(0.3)^2\over2} \right)=4(0.3-0.045)\gt4(0.3-0.05)=1$$
A: It amounts to showing that for $n>2$, $n>\mathrm e$. Now $\mathrm e$ is the limit of the increasing sequence $\;a_n=\Bigl(1+\dfrac1n\Bigr)^n$, and $a_1=2\mkern1.5mu$…
