# 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.

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Using the binomial theorem for integer exponents:

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

Take the nth root, to give:

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

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oh, that's really nice. how to you prove that $(1+\frac{1}{n})^n>2$? Thanks! – Anonymous Mar 28 '12 at 20:18
$(1+x)^n = 1+nx+$ other positive terms – Mark Bennet Mar 28 '12 at 20:20
And $n$ has to be greater than 1. – Mark Bennet Mar 28 '12 at 20:21
sweet, thank you. – Anonymous Mar 28 '12 at 20:25
Now look at Sivaram Ambikasaram's answer and catch hold of how the two are related - then you'll have learned some maths. – Mark Bennet Mar 28 '12 at 20:33

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

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Thank you for the help, however I don't want to prove this using ln and e. – Anonymous Mar 28 '12 at 20:09
@Anonymous: Thank you for your comment; perhaps next time you can state, in your post, what it is you do and do not want, so that other people may not waste their time giving you information you don't care about. – Arturo Magidin Mar 28 '12 at 20:11
I really appreciate your time and sorry for the wasted time for me, however there are probably people who would find this informative and want to know how to prove it this way. I didn't state that I don't want lan and e because I wasn't aware that it can be proven this way. Thank you very much again. – Anonymous Mar 28 '12 at 20:16
This answer deserves more than one upvote, as it not only answers the question, but also deals neatly with the ambiguity in the way the question is asked. – Mark Bennet Mar 28 '12 at 21:37

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 $2^{1/n} < 1 + \epsilon$.

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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: http://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.

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Sorry, had some free time :-) I am pretty sure there are more proofs... – Aryabhata Mar 28 '12 at 21:31
Very nice list! – Pedro Tamaroff Mar 29 '12 at 23:50
@PeterT.off: Thanks! – Aryabhata Mar 29 '12 at 23:59