Prove the truth of $\sqrt[n]{n} - 1 < \frac{1}{\log(n)}$ when $n>1$

Question

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.

Answer 1

Disclaimer: this proof may be unnecessarily complicated, I would be happy to read any simplification or other idea.

We want to prove $$n^\frac{1}{n} < 1+ \frac{1}{\log(n)}.$$

Let us elevate both terms to the positive power $\log(n)$, this reduces to: $$e^\frac{\log(n)^2}{n} < \left(1+ \frac{1}{\log(n)} \right)^{\log(n)}.$$

Observe that the right hand side, which is monotone increasing, goes to $e$ for $n\rightarrow \infty$, but the left hand side, which is monotone decreasing for $n \geq 8$, goes to $1$. Since at $n=8$ the inequality holds we are left with only a finite number of cases to check, where the inequality holds.

Answer 2

Consider $g(y) = e^y - y - y^2$. $g'(y) = e^y - 1-2y$, $\ g''(y) = e^y - 2 > 0 $ for $y > \log 2$ and $g''(y) < 0$ for $y < \log 2$. That is, $g'$ is strictly increasing for $y > \log 2$ and strictly decreasing for $y < \log 2$.

Now, $g'(0) = 0 $, so $g' < 0 $ in $(0,y_0)$ and $g' > 0 $ in $(y_0,\infty)$ for some $y_0 > \log 2$, and $g'(y_0) = 0$. $y_0 > \log 2$ because $g'(\log 2) < 0$. So, $g$ attains its minimum value in $(0,\infty)$ at $y_0$.

Also, $y_0 \in (1,3/2)$, since $g'(1) = e-3 < 0$ and $g'(3/2) = e^{3/2} - 4 >0$. We also have $e^{y_0}=1+2y_0$. Then, $g(y_0) = 1+2y_0 - y_0 - y_0^2 = 1+y_0-y_0^2 > 0$, since $1+y-y^2 > 0$ in $(1,(\sqrt{5}+1)/2) \supset (1,3/2) \ni y_0$. Thus $g(y) \geq g(y_0) > 0$ for $y > 0$.

Now, by the Mean Value Theorem, we have $\log(1+1/y) = \log (y+1) - \log(y) = 1/\eta$ for $\eta \in (y,y+1)$. So, $1/\eta > 1/(y+1) > y/e^y$, since $g > 0$.

Therefore, for $y > 0$, $$\log(1+1/y) > \frac{y}{e^y}$$ Putting $y = \log x$, we have $\log(1+1/\log x) > \log x/x$, that is, $$1+\frac{1}{\log x} > x^{1/x} \text{ for } x >1.$$

Answer 3

Maybe not the best answer but here are a few thought : $$ \begin{align} n^{\frac1n} &< \frac{1}{\ln(n)} + 1 &\overset{(*)}{\Longleftrightarrow} \\ \frac{\ln(n)}{n} &< \ln \left( \frac{1}{\ln(n)} + 1 \right) &\overset{(**)}{\Longleftarrow} \\ \frac{\ln(n)}{n} &< \frac{\frac{1}{\ln(n)}}{\frac{1}{\ln(n)}+1} = \frac{1}{\ln(n) + 1} &\Longleftrightarrow \\ \ln(n)^2 + \ln(n) &< n &\overset{(***)}{\Longleftarrow} \\ 2\ln(n)^2 &< n &\Longleftrightarrow \\ n &< \exp\left( \sqrt{n/2} \right) = \sum_{i=0}^\infty \frac{n^{i/2}}{2^{i/2}\ i!} &\overset{(****)}{\Longleftarrow} \\ n &< \frac{n^3}{5760} &\Longleftrightarrow \\ 75.89 &\approx \sqrt{5760}<n \end{align} $$

$(*)$ Taking $\ln(\cdot)$

$(**)$ Since $\ln(x) \geq \frac{x-1}{x}$

$(***)$ Assuming $n \geq e$

$(****)$ Retaining only the term $i=6$.

Then you "only" need to check that this holds for $n \leq 75$

Answer 4

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} $