Is the infinite root of any number equal to $1$?

Solution 1:

Perhaps the easiest way to see (assuming that $a>0$) that $\sqrt[n]a$ converges to $1$ as $n\to\infty$ is to look at the logarithm: $$\log\sqrt[n]a=\log a^{1/n}=\frac1n\log a\;.$$ By taking $n$ large enough, you can clearly make $\frac1n$ as small as you want, so you can also make $\frac1n\log a$ as small as you want: for a fixed $a>0$, $\log a$ is just some fixed real number, some constant.

Thus, the logarithm of $\sqrt[n]a$ approaches $0$ as $n$ increases. But this means that $\sqrt[n]a$ itself must be approaching $1$, the only numbers whose logarithm is $0$.

(By the way, it doesn’t actually make sense to talk about the ‘infinite root’; you really do have to talk about convergence or limits, as you did in the question proper.)

Solution 2:

Write $a = 1 + b$. If you remember the binomial theorem, $(1 + b/n)^n = 1 + b + $ some other positive terms. So the $n$th root of $1 + b$ is going to be less than $1 + b/n$. But since $1^n = 1 < 1 + b$ you have that the $n$th root of $1 + b$ is greater than $1$. So you have $$1 < (1 + b)^{1 \over n} < 1 + b/n$$ As $n$ goes to infinity, $1 + b/n$ goes to $1$, so by the squeeze theorem you have $$\lim_{n \rightarrow \infty} (1 + b)^{1 \over n} = 1$$

Solution 3:

I use the following theorems:

If $\{ x_n \}_{n \in \Bbb N}$ is a monotonically increasing (decreasing) sequence of real numbers $\Bbb R$ and it is bounded from above (below) then $\{x_n\}_{n \in \Bbb N}$ converges.

We define $x_n = a^{1/n}$ We have that $a>0$. We need to analyze two cases $a<1$ and $a>1$.

For the sake of brevity I'll work simultaneusly on both cases:

$$ a \text{ }{> \choose <}\text{ }1$$

We raise to the power of $n$.

$$ { a }^{\frac 1 n}=x_n \text{ }{> \choose <}\text{ }1$$

This means that if $a>1$ the sequence is bounded from below by $1$ and if $a<1$ it is bounded from above by $1$.

The next step is to prove that

$$x_{n+1} \text{ }{< \choose >}\text{ }x_n$$

This is not rigorous, but note that

$${a^{{1 \over {n + 1}}}} \text{ }{< \choose >}\text{ }{a^{{1 \over n}}} \Leftrightarrow {a^{{n \over {n + 1}}}} \text{ }{< \choose >}\text{ } a \Leftrightarrow {a^{1 - {1 \over n}}} > a \Leftrightarrow {a^{ - {1 \over n}}} \text{ }{< \choose >}\text{ } 1 \Leftrightarrow {a^{{1 \over n}}} \text{ }{> \choose <}\text{ } 1 \Leftrightarrow a \text{ }{> \choose <}\text{ } 1$$

So it is the fact that $a<1$ or $a>1$ that yields all the necessary results. Now we know the sequence converges, so all it's subsequences are convergent. Thus

$$\eqalign{ & \mathop {\lim }\limits_{n \to + \infty } {x_n} = \mathop {\lim }\limits_{n \to + \infty } {x_{2n}} \cr & \mathop {\lim }\limits_{n \to + \infty } {a^{{1 \over n}}} = \mathop {\lim }\limits_{n \to + \infty } {a^{{1 \over {2n}}}} \cr & \mathop {\lim }\limits_{n \to + \infty } {a^{{1 \over n}}} = {\left( {\mathop {\lim }\limits_{n \to + \infty } {a^{{1 \over n}}}} \right)^{{1 \over 2}}} \cr & L = {L^{{1 \over 2}}} \cr} $$

From this it is the case either $L=0$ or $L=1$. But it can't be $L=0$ (why?), so it is $L=1$.