Prove inequality $\frac{1}{2n} < \sqrt[n]{2} - 1 \leqslant \frac{1}{n}$

Question

I tried to solve left side with induction. Assuming that $\frac{1}{2n} < \sqrt[n]{2} - 1 $, we have $\frac{1}{\sqrt[n]{2} - 1} + 2 < 2n + 2$ and $\frac{1}{\sqrt[n]{2} - 1} + 2 > \frac{1}{\sqrt[n+1]{2} - 1}$ is true, so $\frac{1}{2n} < \sqrt[n]{2} - 1 $ is true for every n. But I don't know how to do right side, it doesn't work the same way.

Answer 1

We have that

$$\sqrt[n]{2} - 1 \leqslant \frac{1}{n} \iff n(1+1)^\frac1n - n \leqslant 1$$

and by Bernoulli inequality we have

$$ n(1+1)^\frac1n - n \le n\left(1+\frac1n\right)-n =1$$

Answer 2

Assume the contrary and that $n\ge 1$: $$\sqrt[n]2-1>\frac1n$$ Then $$2>\left(1+\frac1n\right)^n\ge1+\binom n1\frac1n=2$$

Answer 3

hint: $$ 1 \le \root n \of 2 \le 1 + {1 \over n}\quad \left| {\;1 \le n} \right.\quad \Leftrightarrow \quad 2 \le \left( {1 + {1 \over n}} \right)^{\,n} $$ and the RHS is known to be increasing

Answer 4

A very elementary way uses

• $a^n - 1 = (a-1)(a^{n-1} + \cdots + a + 1) \Leftrightarrow a-1 = \frac{a^n - 1}{a^{n-1} + \cdots + a + 1}$ with $\boxed{a=\sqrt[n]{2}}$ and note that
• $ \sqrt[n]{2} >1 \Rightarrow \left(\sqrt[n]{2}\right)^k < 2 = \left(\sqrt[n]{2}\right)^n$ for $k=0,\ldots , n-1$

Hence,

$$\sqrt[n]{2}-1 =\frac{1}{\left(\sqrt[n]{2}\right)^{n-1} + \cdots \sqrt[n]{2} + 1} < \frac{1}{\underbrace{1+ \cdots 1+1}_{n \times 1}} = \frac 1n$$

and $$\frac{1}{\left(\sqrt[n]{2}\right)^{n-1} + \cdots \sqrt[n]{2} + 1} >\frac{1}{\underbrace{2+ \cdots 2+2}_{n \times 2}} = \frac{1}{2n}$$

\begin{eqnarray*} \end{eqnarray*}

Answer 5

The right inequality can be shown from a version of Bernoulli's inequality,

which states that $(1+x)^{r}\leq 1+rx$ for $0 ≤ r ≤ 1 $ and real number $x ≥ -1$,

so, for $n\ge1$, $\sqrt[n]2=(1+1)^{1/n}\le1+\frac1n$.

Answer 6

The sequence $\left(1+\dfrac1n\right)^n$ is growing and known to tend to $e$, hence

$$2\le\left(1+\frac1{n}\right)^n<4$$

and

$$\sqrt[n]2-1\le\frac1{n}<\sqrt[n]4-1.$$

Then right inequality can be written

$$\frac1{2n}<\sqrt[2n]4-1=\sqrt[n]2-1.$$