# Limit of $\frac{1}{\sqrt[n] n}$

I would like to show that the limit of the sequence $$(\frac{1}{\sqrt[n] n})_{n=1}^{\infty}$$ is equal to 1 using the definition of the limit of a sequence. (Another way to say this is that it converges to 1).

Definition: A sequence $$(a_n)_{n=1}^{\infty}$$ converges to a real number A iff for each $$\epsilon > 0$$ there is a positive integer N such that for all $$n \geq N$$ we have $$|a_n - A| < \epsilon$$.

Edit: Here is my work.

Let $$\epsilon > 0$$. There exists $$N$$ such that for $$n \geq N$$,

$$|\frac{1}{\sqrt[n] n}-1| = |\frac{n^{\frac{n-1}{n}}}{n} - 1| = |\frac{n^{\frac{n-1}{n}} - n}{n}| = \frac{n-n^{\frac{n-1}{n}}}{n}$$

Here is where I am stuck.

• In your course, are you allowed to assume that / have you shown that $\exp(x)$ is continuous, and have you shown that continuous functions applied to convergent sequences yield convergent sequences? Nov 24, 2020 at 15:56
• Yes, in fact that's already what I am doing with this. I had the function $f(x) = x^x$ and used the sequence $(x_n) = \frac 1n$ to try and prove that the limit is 1. Nov 24, 2020 at 16:23
• I will point out that you can find several posts related to this limit. For example: How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$? and other questions linked there. Nov 24, 2020 at 18:55

Let $$1-\frac{1}{\sqrt[n]{n}}=a_n. \tag{1}$$ Clearly $$a_n\in(0,1)$$. From (1), one has $$\frac{1}{\sqrt[n]{n}}=1-a_n.$$ So $$n=\frac1{(1-a_n)^n}=(1+a_n+a_n^2+\cdots)^n\ge(1+a_n)^n\ge\binom{n}{2}a_n^2$$ from which one has $$0 Thus, for $$\forall \epsilon>0$$, letting $$\sqrt{\frac{2}{n-1}}<\epsilon$$ gives $$n>\frac{2}{\epsilon^2}+1$$. Define $$N=\lfloor\frac{2}{\epsilon^2}+1\rfloor+1$$ and then when $$n\ge N$$, one has $$0 or $$\left|\frac1{\sqrt[n]n}-1\right|<\epsilon,$$ namely $$\lim_{n\to\infty}\frac1{\sqrt[n]n}=1.$$

• +1 Beat me to it. You should have an $\epsilon^2$ in some of your denominators. Nov 24, 2020 at 16:23
• Thanks, just forgot to type the square. Nov 24, 2020 at 16:31

An old standby:

By Bernoulli's inequality, $$(1+n^{-1/2})^n \ge 1+n^{1/2} \gt n^{1/2}$$.

Raising to the $$2/n$$ power, $$n^{1/n} \lt (1+n^{-1/2})^2 =1+2n^{-1/2}+n^{-1} \lt 1+3n^{-1/2} \to 1$$.

Not sure if that would help

your equation $$= \dfrac{1}{n^\frac{1}{n}} = n^\frac{-1}{n} = \exp\left(\dfrac{-1}{n}*\ln(n)\right)$$

$$\dfrac{-1}{n}*\ln(n)$$ goes to $$0$$ when $$N$$ goes to $$\infty$$

Therefore $$\exp(0) = 1$$

• It's best that you mention Stolz–Cesàro: $\lim_{n\to\infty} \frac{\ln(n)}{n} = \lim_{n\to\infty} \frac{\ln(n/(n-1))}{n-(n-1)}=0$. Nov 24, 2020 at 16:47
• @NeatMath new name to add to my list.
– ombk
Nov 24, 2020 at 16:50

Let's do this just using straightforward algebra and some crude (but clever) bounds.

For $$n\ge1$$ we have, using the algebraic identity $$(1-a^n)=(1-a)(1+a+a^2+\cdots+a^{n-1})$$ and the inequality $$\sqrt[n]n\ge1$$,

\begin{align} 0\le1-{1\over\sqrt[n]n}&=1-n^{-1/n}\\ &=(1-n^{-1/n}){1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}\over1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}}\\ &={1-n^{-1}\over1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}}\\ &\lt{1\over(n^{-1/2}+n^{-1/2}+\cdots+n^{-1/2})+(0+0+\cdots+0)}\qquad\text{(see note below)}\\ &\lt{1\over(n/3)n^{-1/2}}\\ &={3\over\sqrt n}\\ &\lt\epsilon\qquad\text{if n\gt{9\over\epsilon^2}} \end{align}

The key step is the replacement of the (roughly) first third of the terms in $$1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}$$ with the smaller value $$n^{-1/2}$$ and the rest with $$0$$; one can, of course, go up close to the first half and replace the $$3$$ here with something close to $$2$$, but there's not much to be gained in doing so.

Pick $$\varepsilon>0$$. We have

$$\sqrt[n]{n}-1 < \varepsilon \iff \sqrt[n]{n} < 1+\varepsilon \iff n < (1+\varepsilon)^n$$

Now, we have $$(1+\varepsilon)^n = 1+n\varepsilon + {n \choose 2}\varepsilon^2 + \cdots > {n \choose 2}\varepsilon^2 = \frac12(n^2-n)\varepsilon^2$$

so if we pick $$n \ge \sqrt{1+\frac2{\varepsilon^2}} \implies n \le \frac12(n^2-n)\varepsilon^2 < (1+\varepsilon)^n$$

and hence $$\lim_{n\to\infty} \sqrt[n]{n} = 1$$. Now therefore also $$\lim_{n\to\infty} \frac1{\sqrt[n]{n}} = 1.$$

I think the following method can help much. For each $$n$$, there exists an integer $$k$$ such that $$2^{k-1}\le n\le2^k$$ and we have either $$2^{k-1\over 2^{k-1}}\le n^{1\over n}\le 2^{k\over 2^k}$$ or $$2^{k\over 2^{k}}\le n^{1\over n}\le 2^{k-1\over 2^{k-1}}$$ Now prove $$k\over 2^k$$ tends to $$0$$.