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if $a>1$, Prove that $\lim a^{1\over n}=1$ Is the result true if $0<a\le1 ?$

My attempt :

let $a^{1/n}=1+h$, then $a=1+nh+\frac{n(n-1)h^2}{2}+\dots+h^n$

so, $a>\frac{n(n-1)h^2}{2}$

or, $|h|<\sqrt{\frac{2a}{n}}$

for given $\epsilon>0$

$|a^{1/n}-1|=|h|<\sqrt{\frac{2a}{n}}<\epsilon$

i.e $n>\frac{2a}{\epsilon^2}$

Hence we can find a suitable $m$ for all $a>0$

Is it correct ?

Aman Mittal
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2 Answers2

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Your proof looks fine.

This is how I would prove it.

If $a>1$, put $x_n=\sqrt[n]{a}-1$. Then $x_n>0$, and, by the binomial theorem, $$1+nx_n\le \left(1+x_n\right)^n=a$$ so that $$0<x_n\le \frac{a-1}{n}$$ Hence $x_n\rightarrow 0$. If $a=1$ this is trivial, and if $0<a<1$, the result is obtained by taking recipocals!

hrkrshnn
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  • Can you show this $0<a<1$ case? Taking reciprocals I'm getting $0>a_n \ge \frac{n}{a-1}$ what for $0<a<1$ only means $0>a_n \ge -\infty$ – Piotr Wasilewicz Mar 13 '24 at 09:50
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Apart from fact that we need to specify that $n\ge 2$, the argument is correct for $a\gt 1$. It would be simpler to use the fact that if $n\ge 1$ then $a\ge 1+nh$.

However, the conclusion that we can find a suitable $m$ for all $a\gt 0$ has not been proved. For the argument, though correct when $h$ is positive, does not work if $h\lt 0$. For the case $0\lt a\lt 1$, the standard thing to do is to let $b=\frac{1}{a}$. Because the sequence $(b^{1/n})$ converges to $1$, so does the sequence $(a^{1/n})$.

André Nicolas
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  • why would the conclusion not be true for $h<1 ?$ – Aman Mittal Jul 09 '14 at 17:21
  • The answer above does not say for $h\lt 1$, it is for $a\lt 1$, meaning $h\lt 0$. Then the terms of the binomial expansion alternate in sign, so without further analysis we cannot say anything about the size of the truncation error. Now in fact we do have convergence to $1$ for $0\lt a\lt 1$, but we need a different proof for that case. – André Nicolas Jul 09 '14 at 17:32