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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.

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    $\begingroup$ 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? $\endgroup$ Commented Nov 24, 2020 at 15:56
  • $\begingroup$ 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. $\endgroup$
    – o's1234
    Commented Nov 24, 2020 at 16:23
  • $\begingroup$ 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. $\endgroup$ Commented Nov 24, 2020 at 18:55

6 Answers 6

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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 $$ 0<a_n\le\sqrt{\frac{2}{n-1}}. $$ 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 $0<a_n<\epsilon$ or $$ \left|\frac1{\sqrt[n]n}-1\right|<\epsilon,$$ namely $$ \lim_{n\to\infty}\frac1{\sqrt[n]n}=1. $$

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

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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 $\exp(0) = 1$

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

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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 $$\lim_{n\to\infty} \frac1{\sqrt[n]{n}} = 1.$$

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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 $0$.

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