Let $x_{n} = (n)^{\frac{1}{n}} - 1.$ Use the fact that $(1+x_{n})^n = n$ To show that $(x_{n})^2 \leq \frac{2}{n}$

Question

Let $$x_{n} = (n)^{\frac{1}{n}} - 1.$$

Use the fact that $$(1+x_{n})^n = n$$

To show that $$(x_{n})^2 \leq \frac{2}{n}$$

Hint: Use Binomial Theorem and trow away most terms

Attempt:

I've tried all sorts of manipulations, but I stil feel I am at ground zero.

I first expanded $$(x^{2}_{n}) = n^{\frac{2}{n}} - 2n^{\frac{1}{n}} + 1$$

So I feel that the next thing that is true is that:

$$(x^{2}_{n}) = n^{\frac{2}{n}} - 2n^{\frac{1}{n}} + 1 \leq n^{\frac{2}{n}} + 2n^{\frac{1}{n}} + 1 = (n^{\frac{1}{n}} + 1)^{2} = (x_n + 1 + 1)^{2} = (x_n + 2)^2 $$

There is some trick that I am missing that would get me to my conclusion but I can't seem to figure it out. Hints?

Answer 1

Use the fact that $(1+x_{n})^n = n \;\;\ldots\;\;$ Hint: Use Binomial Theorem and trow away most terms

Using the given hint, and discarding all (positive) terms except the first and third ones:

$$\require{cancel} \begin{align} 1+ \binom{n}{1}x_n+\binom{n}{2}x_n^2+ \ldots = n \;\;&\implies\;\; 1 + \frac{n(n-1)}{2}x_n^2 \le n \\ &\iff\;\; n\bcancel{(n-1)}x_n^2 \le 2\bcancel{(n-1)} \end{align} $$

Answer 2

Use $(1+x_n)^n=\sum_{i=0}^n{n \choose i}( x_n)^{n-i}$ by the binomial theorem

So $1+nx_n+\frac {n(n-1)}2(x_n)^2+...=n \implies \frac {n(n-1)}2(x_n)^2\le n-1 \implies (x_n)^2\le \frac2n$

Answer 3

An alternative approach: $1\cdot\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)=n$, hence by the AM-GM inequality

$$\begin{eqnarray*} n^{1/n}&=&\text{GM}\left(1,\frac{3}{2},\frac{4}{3},\ldots,\frac{n}{n-1}\right)\\&\leq&\text{AM}\left(1,\frac{3}{2},\frac{4}{3},\ldots,\frac{n}{n-1}\right)\\&=&\frac{1}{n}\left(1+\frac{3}{2}+\frac{4}{3}+\ldots+\frac{n}{n-1}\right)\\&=&\frac{n-1+H_{n-1}}{n}\end{eqnarray*} $$ and $n^{1/n}-1\leq \frac{H_{n-1}-1}{n}$ for any $n\geq 2$. In order to prove the claim, it is enough to show $(H_{n-1}-1)^2\leq 2n$. On the other hand, by the Cauchy-Schwarz inequality

$$ (H_{n-1}-1)^2 = \left(\sum_{k=2}^{n-1}\frac{1}{k}\right)^2 \leq \sum_{k=2}^{n-1}1\sum_{k=2}^{n-1}\frac{1}{k^2}\leq (n-2)(\zeta(2)-1) $$ hence the inequality we want to prove is pretty loose.