Limit of $2^{1/n}$ as $n\to\infty$ is 1

How do I prove that:

$\lim \limits_{n\to \infty}2^{1/n}=1$

Thank you very much.

Solution 1:

Using the binomial theorem for integer exponents:

Can you see that $(1+\frac 1 n)^n > 2>1$

Take the nth root, to give:

$(1+\frac 1 n) > 2^{\frac 1 n} >1$

Solution 2:

Note that for $a\gt 0$, $$a^b = e^{b\ln a}.$$ So $$\lim_{n\to\infty}2^{1/n} = \lim_{n\to\infty}e^{\frac{1}{n}\ln 2}.$$ Since the exponential is continuous, we have $$\lim_{n\to\infty}e^{\frac{1}{n}\ln 2} = e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln 2}.$$

Can you compute $\displaystyle\lim_{n\to\infty}\frac{\ln 2}{n}$ ?

Solution 3:

HINT: Use squeeze theorem.

Since $1 < 2$, we have $1 = 1^{1/n} < 2^{1/n}$ for all $n \in \mathbb{N}$.

To bound the limit from above, note that $1 + n \epsilon < \left( 1 + \epsilon \right)^n$.

Hence, given any $\epsilon >0$, $\forall n > \displaystyle 1/{\epsilon}$, we have $2 < 1 + n \epsilon < \left(1 + \epsilon \right)^n$ and hence $2^{1/n} < 1 + \epsilon$.

Solution 4:

Here are a few different proofs which don't use $e$ or $\log$ and can be regarded completely elementary.

Proof 1)

We use the following theorem:

If $a_n \gt 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim_{n\to\infty} a_n^{1/n} = L$

This is a standard theorem, and a proof can be found in almost any textbook. You can also find a proof in my answer here: Show that this limit is equal to $\liminf a_{n}^{1/n}$ for positive terms.

Apply the theorem to the sequence $a_n = 2$.

Proof 2)

We use the $\text{AM} \ge \text{GM}$ inequality on $n-1$ ones and one $2$.

$$\frac{1 + 1 + \dots + 1 + 2}{n} \ge 2^{1/n}$$

$$ \frac{n+1}{n} \ge 2^{1/n}$$

Thus we have

$$ 1 + \frac{1}{n} \ge 2^{1/n} \ge 1$$

so by Squeeze theorem, $\lim_{n \to \infty} 2^{1/n} = 1$.

Proof 3)

We can use Bernouli's inequality (essentially similar to Sivaram's answer) to show that

$$\left(1 + \frac{1}{n}\right)^n \ge 1 + \frac{1}{n} \times n = 2$$

and we get inequalities similar to the proof in 2).

Proof 4)

The sequence $a_n = 2^{1/n}$ is bounded below (by $1$) and montonically decreasing.

Thus it is convergent, to say $L$. Since $a_{2n}$ also converges to $L$, we have that $L = \sqrt{L}$, as $2^{1/2n} = \sqrt{2^{1/n}}$. So $L = 0$ or $L = 1$. Since the limit is not less than $1$ ($2^{1/n} \ge 1$), the limit is $1$.

Proof 5)

For $n \gt 2$, we have that $1 \le 2^{1/n} \le n^{1/n}$.

Now use the fact that $\lim_{n \to \infty} n^{1/n} = 1$.

An elementary proof of that can be found here: https://math.stackexchange.com/a/115825/1102. Any proof for $n^{1/n}$ now becomes a proof for $2^{1/n}$. Proof 1) above can also be used for $n^{1/n}$.

Proof 6)

Using combinatorics.

The number of $n$ digit numbers in base-$n+1$ is $(n+1)^n$ (allowing for leading zeroes). The number of $n$ digits numbers in base-$n$ is $n^n$. We can show that $(n+1)^n \ge 2 \times n^n$: consider the base-$n$ numbers. Replace the last digit with $n$. You get a base-$n+1$ $n$ digit number. Counting the base-$n$ numbers (which are also base-$n+1$ numbers) and the "last digit modified" numbers, gives us the inequality.

This inequality implies that $1 + \frac{1}{n} \ge 2^{1/n}$ and can be used to give a proof using the squeeze theorem, similar to proofs 2 and 3.