<s>[INST] Prove the truth of $\sqrt[n]{n} - 1 < \frac{1}{\log(n)}$ when $n>1$ part of a proof I'm writing for homework I came across the following inequality and I feel like I don't have the right tools to solve it.
$$ \sqrt[n]{n} - 1 < \frac{1}{\log(n)}, 1<n\in \mathbb{N}$$
I thought to solve it by induction, but every time I'm getting stuck at simplifying the expression.
 [/INST] To show
$\sqrt[n]{n} - 1 < \frac{1}{\log(n)}, 1<n\in \mathbb{N}
$,
the inequalities needed are
$\ln(2) < .7$,
$\frac{\ln(x)}{x}
\le \frac1{e}
$
(since $\frac{\ln(x)}{x}$
has its maximum at $e$),
$\ln(x) \le x-1
$
for
$x \ge 1$
(compare derivatives),
and
$e^x \le \frac1{1-x}$
for $0 < x < 1$
(compare coefficients in 
the power series).
We have
$n^{1/n}
= e^{\ln(n)/n}
\le \frac1{1-\ln(n)/n}
$,
so we are done if
we can show that
$ \frac1{1-\ln(n)/n}
\le 1+\frac{1}{\ln(n)}
$.
This is equivalent to
$1 \le 1+\frac{1}{\ln(n)}-\frac{\ln(n)}{n}-\frac1{n}
$
or
$\frac{\ln(n)+1}{n}
\le \frac{1}{\ln(n)}
$
or
$\ln^2(n)+\ln(n) \le n$.
We can ask Wolfy,
who says that
$f(x)
=x-\ln^2(x)-\ln(x)
$
has a root at
$x=0=0.290734083070702...$
and is positive for
$x > x_0$.
To prove it,
I will show that
$f(x) > 0$ for $x \ge 4$;
computing $f(1), f(2),$
and
$f(3)$ will complete the proof.
$f(1) = 0,
f(2) = 0.826...,
$
and
$f(3) = 0.694...$.
$f'(x)
= \frac{x - 2 \ln(x) - 1}{x}
$
so we are done if
$x - 2 \ln(x) - 1
\ge 0$.
If
$g(x) = x - 2 \ln(x) - 1$,
then
$g'(x) = 1-\frac{2}{x}
\gt 0$
for $x \gt 2$
and
$g(4)
=3-4\ln(2)
\gt 3-4\cdot .7
\gt 0
$
Therefore
$g(x) > 0$
for
$x \ge 4$.
Going back to $f(x)$,
$f'(x) \gt 0$
for $x \ge 4$
and
$\begin{array}\\
f(4)
&=4-\ln^2(4)-\ln(4)\\
&=4-2\ln(2)(2\ln(2)+1)\\
&\gt 4-1.4\cdot 2.4\\
&=0.34\\
&> 0\\
\end{array}
$
</s>