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Using $$(a+b)^{n} = \sum_{i=0}^n {n \choose k} a^{n-k} b^{K}$$

prove that

$$n^{1/n} - 1 \le \sqrt{\frac 2n}$$ for n= 2,3,4....

I know the first step is to set $$ n^{\frac 1n} = 1 + x $$ for some x>0 and then raise both sides to the n, but I'm lost after that.

Martin Sleziak
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Ryan T. Donnelly
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3 Answers3

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If $x=n^{1/n}-1$, then $n=(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\cdots\ge 1+\frac{n(n-1)}{2} x^2$ for $n\ge 2$.

Then $n-1\ge \frac{n(n-1)}{2} x^2\implies 1\ge \frac{n}{2}x^2\implies x^2\le\frac{2}{n}\implies x\le\sqrt{\frac{2}{n}}$

user84413
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Let's look at $(1+\sqrt\frac{2}{n})^n=1+n\sqrt\frac{2}{n}+\frac{n(n-1)}{2}\frac{2}{n}+\text{more positive terms}= 1+(n-1)+n\sqrt\frac{2}{n}+...\geq n$, now take the $n^{th}$ root of both sides and you get the desired results.

Two remarks: This implicitly assumes $n>2$, but you can simply check for $n<2$. I did not use induction, are you required to use induction or was it your guess that this is the way to approach this problem?

Shahar Even-Dar Mandel
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Here is a better result, gotten by elementary means.

By Bernoulli's inequality, $(1+n^{-1+1/k})^n \ge 1+n^{1/k} > n^{1/k} $.

Raising to the $k/n$ power, $n^{1/n} < (1+n^{-1+1/k})^k $.

If $k = 3$, this bound is $(1+n^{-2/3})^3 =1+3n^{-2/3}+3n^{-4/3}+n^{-2} <1+7n^{-2/3} $.

Extending this, I have shown that $n^{1/n} < 1+2kn^{–1+1/k}$ for $n>k^{k/(k-1)}$.

marty cohen
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