$a_n= \sqrt[n]{n}-1$ , show $a_n ^2< \frac{2}{n-1}$

Question

Let $a_n= \sqrt[n]{n}-1$ . Show that $a_n ^2< \frac{2}{n-1}$ for $n>1$, where $n$ is a natural number.

Can anybody please help solving this problem?

Answer 1

We need to prove that $$\left(\sqrt[n]n-1\right)^2<\frac{2}{n-1}$$ or $$n<\left(1+\sqrt{\frac{2}{n-1}}\right)^n,$$ which is true because for all natural $n\geq2$ by the binomial of Newton we obtain: $$\left(1+\sqrt{\frac{2}{n-1}}\right)^n\geq1+n\cdot\sqrt{\frac{2}{n-1}}+\frac{n(n-1)}{2}\cdot\frac{2}{n-1}>n$$

The binomial of Newton it's the following.

For all natural $n\geq2$ we obtain: $$(a+b)^n=a^n+na^{n-1}b+\frac{n(n-1)}{2}a^{n-2}b^2+,,,+b^n.$$ In our case $a=1$ and $b=\sqrt{\frac{2}{n-1}}.$