It makes sense to me that for any $p>1$, the nth root will decrease towards $1$ and for any $p<1$, the nth root will increase towards 1 and this back-and-forth nature will make the sequence converge to $1$. I just cannot think of any way to show it. Even a nudge in the right direction will be greatly appreciated.
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What does "back and forth" mean here? One approach is to look at logarithms – Henry Sep 20 '22 at 16:54
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3Does this answer your question? Understanding an MSE proof of $\lim_{n \to \infty} p^{\frac{1}{n}} = 1$ ($p >0$) – Dietrich Burde Sep 20 '22 at 16:57
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- For any $a<1$ then $a^n\to 0\quad$ (a positive)
- For any $b>1$ then $b^n\to +\infty$
$p$ being fixed, for $n$ large enough we will have $a^n<p<b^n$ (there is plenty of room).
This implies that $a<\sqrt[n]{p}<b$, and since $a,b$ are arbitrary, this squeezes the limit to be exactly $1$ (i.e. take $a=1-\epsilon,\ b=1+\epsilon$).
zwim
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Formally, you can do a $\varepsilon$ proof for this. Choose arbitrarily small $\varepsilon>0$. Now, try to find some (large) $N\in \mathbb N$ such that for every $n>N$, you'll have $|\sqrt[n]{p}-1|<\varepsilon$, so if $p>1$, this implies $\sqrt[n]{p}<\varepsilon+1$. How big does $N$ have to be?
Logan Post
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