What is the limit of $3^{1/n}$ when n approaches infinity

Question

Graphically, I see that $\lim_{n->\infty}3^{1/n}$ approaches $1$. However, how to show $\lim_{n->\infty}3^{1/n} = 1$ step by step?

Answer 1

HINT:

Using $e^x\le \frac1{1-x}$ for $x<1$, we have for $n\ge 2$

$$1< 3^{1/n}< \frac{1}{1-\frac1n \log(3)}$$

Now apply the squeeze theorem.

Answer 2

Note that

$$\lim_{n->\infty}\frac1n = 0 \quad \quad 3^0=1$$

thus by algebraic rule and continuity

$$\lim_{n->\infty}3^{1/n}=3^{\lim_{n->\infty}{1/n}}$$

we have

$$3^{1/n}\to 3^0=1$$

Answer 3

For any number $a>0$, $a^{1/n}$ tends to $1$ when $n$ tends to infinity since $$\log a^{1/n}=\frac1n\,\log a\to 0\enspace\text{when }\enspace n\to\infty.$$

Answer 4

You only need two elementary results: The binomial theorem and the squeeze theorem.

The binomial theorem implies that if $x>0$ then $(1+x)^n\geq 1+nx$.

Letting $x_n=3^{1/n}-1$, then $$(1+x_n)^{n}=3.$$ Since $x_n>0$, we have, by the above result, that $$3=(1+x_n)^n \geq 1+nx_n.$$

So $0\leq x_n\leq\frac{2}{n}$. Hence $x_n\to 0$ by the squeeze theorem, and and hence $3^{1/n}=x_n+1\to 1.$

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This works for any sequence $a^{1/n}$ with $a>1$ since $x_n=a^{1/n}-1$ gives us that $0<x_n\leq \frac{a-1}{n}.$

To show it for $0<a<1$, we just need that $f(x)=\frac{1}{x}$ is continuous at $x=1.$ Assuming that, we use that $$ a^{1/n}=f\left(\left(a^{-1}\right)^{1/n}\right)$$ And if $0<a<1$, then $1<a^{-1}$ and by our previous result: $$\left(a^{-1}\right)^{1/n}\to 1.$$ So $a^{1/n}\to f(1)=1.$

Answer 5

As an alternative, with a more advanced approach by the ratio-root criteria

$$a_n = \sqrt[n]{b_n} \quad b_n=3$$

$$\frac{b_{n+1}}{b_n} \rightarrow L\implies a_n=b_n^{\frac{1}{n}} \rightarrow L$$

thus since

$$\frac{b_{n+1}}{b_n}=\frac33=1 \implies a_n = \sqrt[n]{3}\to 1$$

Answer 6

Binomial theorem:

For $x\ge 0:$

$(1+x)^n \ge 1 + nx + n(n-1)\dfrac{x^2}{2!} \ge $

$n^2\dfrac{x^2}{4}$ for $n\ge 2.$

Set $x=\dfrac{2√3}{n}:$

$(1+\dfrac{2√3}{n})^n \ge n^2\dfrac{(4)(3)}{4n^2}= 3;$

$ (1+\dfrac{2√3}{n})^{n} \ge 3$ , or

$1 +\dfrac{2√3}{n} \ge 3^{1/n} .$

With lower bound $1$:

$1\lt 3^{1/n} \le 1+ \dfrac{2√3}{n}.$

Limit $n \rightarrow \infty$ is?

Answer 7

Bernoulli's Inequality, which, for integer exponents, can be proven using a simple inductive argument, says $$ \left(1+\frac2n\right)^n\ge3\ge1 $$ Taking roots, we get $$ 1\le3^{1/n}\le1+\frac2n $$ Then, the Squeeze Theorem ensures that $$ \lim_{n\to\infty}3^{1/n}=1 $$