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If $n \geq 2$ is an integer, show

$n^{1/n} = 1 + h$; where $h \leq \sqrt{ \dfrac{2}{n-1}}$

Then Deduce that: $\lim\limits_{n \to \infty} n^{1/n} = 1$

Hint: Since $n>1$, $n^{1/n}>1$. So, $n^{1/n}=1+h$ for some $h>0$. Hence, $n=(1+h)^n$. Then use the binomial theorem to show that

$(1+h)^n \geq \dfrac{n(n-1)}{2} h^2$.

Thus, deduce that $h \leq \sqrt{ \dfrac{2}{n-1}}$

Using the Binomial Theorem:

$$n=(1+h)^n = 1+ nh^1 + \frac{n(n-1)}{2}h^2 + ... + nh^{n-1} + h^n$$

I see that $n \geq\dfrac{n(n-1)}{2}h^2$, it looks very obvious but I don't know how to prove it.

Paramanand Singh
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  • Also you can prove $,n/q^n \rightarrow 0$ when $q>1$ (by the monotone bounded principle) and using this fact to show $1\le n \le (1+\epsilon)^n$ for each $\epsilon >0$. Thus the result is almost trivial. Or shows that if $|x|>1$, then $x^n \rightarrow +\infty$ – Jose Antonio Nov 10 '13 at 03:17

2 Answers2

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We have $h>0$ so $h^{m}>0$ for $m=1,\ldots,n$. Thus,

$$n=(1+h)^{n}=1+nh+\frac{n(n-1)}{2}h^{2}+\cdots+h^{n}>\frac{n(n-1)}{2}h^{2}$$

Pedro
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user71352
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0

Hint:

$$n=1+n\cdot h+{n\choose 2}h+\cdots + {n\choose n-2}h^{n-2}+n\cdot h^{n-1}+h^n$$

You have already identified the required inequality:

$$n\ge \frac {n(n-1)}2h$$

Start by noting that $h\gt 0$ since $n\gt 1$.

abiessu
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