Prove that ln(n) < n for the comparison test I have to prove that for $\ln(n) < n $, while n>0 for the comparison test. I know its intuitive that this would be true, but I am required to actually prove it. It has occurred to me to just put e on both sides, but I didn't think that would be right. Any help is appreciated.
 A: Suppose $x \in (0,1)$, then $\ln x = \int_1^x {1 \over t} dt \le 0$ and
hence $\ln x \le x$.
For $x \ge 1$ we have $\ln x = \int_1^x {1 \over t} dt \le \int_1^x dt = x-1 \le x$.
A: You could show stronger result. Let $\varepsilon > 0$. Then $\ln n < n^\varepsilon$ for all but finitely many $n\in\mathbb N$.
To prove it, note that $$\lim_{x\to\infty}\frac{\ln x}{x^\varepsilon} = 0$$ (by l'Hospital, for example), so there exists $M>0$ such that $n\geq M$ implies $\frac{\ln n}{n^\varepsilon}<1$, i.e. $\ln n<n^\varepsilon$.
Example where this is useful - prove convergence of $\sum_{n=0}^\infty\frac{(\ln n)^2}{n^{3/2}}.$ (Hint: $\frac{(\ln n)^2}{n^{3/2}} < \frac{1}{n^{1+1/8}}$, for all but finitely many $n$.)
A: We have for $n \in \mathbb N$:
$e^n=1+n+\frac{n^2}{2!}+...>n$.
Hence (taking logarithm , observe $\ln$ is strictly increasing)
$n > \ln n$.
A: Hint: for $0<n\leq1$ we know $\ln(n)\leq 0$ , but $n$ is positive. Now take derivatives for $n\geq1$. 
Alternatively, Lord Sharks comment is another straightforward way of going about it.
A: We need to prove that 
$$ f(x) = e^x - x > 0 $$ for all $x \geq 0$. (When $x < 0$, it is obvious.)
Function $f$ is strictly increasing when $x \geq 0$ since:
$$
f'(x) = e^x - 1 \geq 0, \quad \forall \, x \geq 0.
$$ 
Hence for all $x \geq 0$:
$$
f(x) = e^x - x \geq f(0) = e^0 - 0 = 1 > 0 \quad \Longrightarrow
\quad e^x > x
$$
A: By elementary means:
Consider the base-$2$ logarithm, for convenience.
$$n>1\land\text{lg}(n)<n\implies 2n>1\land\text{lg}(n)+1<n+1<2n$$
and by induction and by monotonicity of $\log$ and the identity function, this holds for all $n>1$.
For larger bases, the inequality is strengthened.
