Understanding the proof of $\displaystyle\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$

Question

I need help understanding this:

Prove that $$\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$$

$$(\sqrt n)^\frac{1}{n}\geq (\sqrt1)^\frac{1}{n}=1, \forall n\in\Bbb N$$ By binomial theorem we get for $n\geq 2$

$$n=((\sqrt n)^\frac{1}{n})^n=[1+((\sqrt n)^\frac{1}{n}-1)]^n=\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k$$ $$\geq 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2=1+\frac{n(n-1)}{2}(\sqrt n^\frac{1}{n}-1)^2$$ $$\Rightarrow (\sqrt n^\frac{1}{n}-1)^2\leq \frac{2}{n}$$ $$\Rightarrow \sqrt n^\frac{1}{n}\leq 1+\frac{\sqrt 2}{\sqrt n}$$ $1+\frac{\sqrt 2}{\sqrt n}$ approaches $1$ for $n \rightarrow \infty$ so by the sandwich theorem we get $\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$

I don't understand this part: $$ \sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k\geq 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2.$$ Could someone explain how they got that on the RHS?

Answer 1

Since $n^{1/n}\ge 1$, then all of the terms in the binomial expansion

$$\left(1+(\sqrt[n]{n}-1)\right)^n=\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k$$

are non-negative. Therefore, we have for any $m\le n$

$$\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k\ge \sum_{k=0}^m\binom{n}{k}(\sqrt[n]{n}-1)^k$$

Taking $m=2$ and $n\ge 2$ yields

$$\begin{align} \sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k&\ge \sum_{k=0}^2\binom{n}{k}(\sqrt[n]{n}-1)^k\\\\ &=1+n(\sqrt[n]{n}-1)+\frac{n(n-1)}{2}(\sqrt[n]{n}-1)^2\\\\ &\ge 1+\frac{n(n-1)}{2}(\sqrt[n]{n}-1)^2 \end{align}$$

where the last inequality is true since $n(\sqrt[n]{n}-1)\ge 0$.

Answer 2

They are truncating the binomial expansion, discarding positive terms: if $$ r=r_0+r_1+r_2+\dots+r_n $$ and $r_i\ge0$ for $i=0,1,\dots,n$, then $r\ge r_0+r_2$.

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It can be done more easily. Set $(\sqrt{n})^{1/n}=1+a_n$, so $$ \sqrt{n}=(1+a_n)^{n}\ge1+na_n $$ by Bernoulli’s inequality. Hence $$ a_n\le\frac{\sqrt{n}-1}{n}=\frac{1}{\sqrt{n}}-\frac{1}{n} $$ so $$ (\sqrt{n})^{1/n}=1+a_n\le 1+\frac{1}{\sqrt{n}}-\frac{1}{n} $$ and the squeeze theorem allows to conclude.

Answer 3

Set $l =\lim_{n\to\infty}n^\frac{1}{n}$

$$\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=\sqrt{\lim_{n\to\infty}n^\frac{1}{n}}=\sqrt{l}$$

For the limit of the sequence $n^\frac{1}{n}$ get the sub-sequence $(2n)^\frac{1}{2n}$

So:

$l^2=(\lim (2n)^\frac{1}{2n})^2=\lim (2n)^\frac{1}{n}=\lim 2^\frac{1}{n}. n^\frac{1}{n}=\lim 2^\frac{1}{n}.\lim n^\frac{1}{n}=l$

$l^2=l$ and as $n^\frac{1}{n}\gt1$ we have $l=1$

And $\sqrt{\lim_{n\to\infty}n^\frac{1}{n}}=\sqrt{1}=1$

Answer 4

That proof seems a little complicated; I'll do it a different way. Let's write $(\sqrt n)^{1/n} = 1+a_n.$ We know each $a_n>0.$ For $n\ge 2,$ the binomial theorem shows

$$\sqrt n= (1+a_n)^{n} = 1^n\cdot a_n^0 + n\cdot1^{n-1}\cdot a_n^1 + [n(n-1)/2]\cdot 1^{n-2}\cdot a_n^2 + \cdots > [n(n-1)/2]a_n^2.$$

Therefore

$$\frac{\sqrt n}{n(n-1)/2} > a_n^2.$$

Since the expression on the left $\to 0,a_n^2 \to 0,$ which implies $a_n \to 0.$ It follows that $(\sqrt n)^{1/n} \to 1.$

Answer 5

A Different Approach

Because $\left(1+\frac1n\right)^{n+1}$ is decreasing, for $n\ge3$, $$ \begin{align} \frac{(n+1)^n}{n^{n+1}} &=\frac1{n+1}\left(1+\frac1n\right)^{n+1}\\ &\le\frac14\cdot\frac{256}{81}\\[6pt] &\lt1 \end{align} $$ Thus, for $n\ge3$, $$ (n+1)^{\frac1{n+1}}\lt n^{\frac1n} $$ That is, $n^{\frac1n}$ is decreasing and bounded below by $1$. So $\lim\limits_{n\to\infty}n^{\frac1n}$ exists. $$ \begin{align} \lim_{n\to\infty}n^{\frac1n} &=\lim_{n\to\infty}(2n)^{\frac1{2n}}\\ &=\lim_{n\to\infty}2^{\frac1{2n}}\lim_{n\to\infty}n^{\frac1{2n}}\\ \end{align} $$ Therefore, $$ \begin{align} \lim_{n\to\infty}n^{\frac1{2n}} &=\lim_{n\to\infty}2^{\frac1{2n}}\\[6pt] &=1 \end{align} $$

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Bernoulli's Inequality

Bernoulli's Inequality says $$ \begin{align} \lim_{n\to\infty}\sqrt{n}^{1/n} &=\lim_{n\to\infty}\left(1+\left(\sqrt{n}-1\right)\right)^{1/n}\\[3pt] &\le\lim_{n\to\infty}\left(1+\tfrac1n\left(\sqrt{n}-1\right)\right)\\[6pt] &=1 \end{align} $$ Since $\sqrt{n}^{1/n}\ge1$, the Squeeze Theorem says $$ \lim_{n\to\infty}\sqrt{n}^{1/n}=1 $$

Answer 6

$$\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k= 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2+\left[{{n}\choose {1}}(\sqrt n^\frac{1}{n}-1)^1+\sum_{k=3}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k\right]$$

and see that:

$${{n}\choose {1}}(\sqrt n^\frac{1}{n}-1)^1+\sum_{k=3}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k \ge 0$$

because

$$\sqrt{n}^\frac{1}{n}-1 \ge0$$

Answer 7

I'd like to offer another method, using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$

$$n-1=(n^{\frac{1}{2n}})^{2n}-1=\\(n^{\frac{1}{2n}}-1)(n^{\frac{1}{2n}(2n-1)}+n^{\frac{1}{2n}(2n-2)}+...+n^{\frac{1}{2n}2}+n^{\frac{1}{2n}}+1)\geq$$ using AM-GM $$\geq(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}(2n-1)}\cdot n^{\frac{1}{2n}(2n-2)}\cdot ...\cdot n^{\frac{1}{2n}2}\cdot n^{\frac{1}{2n}}}+1\right)=\\ (n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}(2n-1+2n-2+...+1)}}+1\right)= \\(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}\frac{(2n-1)2n}{2}}}+1\right)=\\ (n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt{n}+1\right)$$ Also $\ln{x}\leq x-1,x>0$ which means $$\ln{n^{\frac{1}{2n}}}=\frac{\ln{n}}{2n}\leq n^{\frac{1}{2n}}-1$$

Altogether $$\frac{\ln{n}}{2n}\leq n^{\frac{1}{2n}}-1\leq \frac{n-1}{(2n-1)\sqrt{n}+1}<\frac{1}{2\sqrt{n}}$$

Answer 8

Let $\lim_{n\to\infty} n^{1/2n} = L$

Taking logarithms, we get ;

$log(L) =$$\lim_{n\to\infty}$$\frac{1}{2n}$$log(n)$

Applying $LH$ Rule :

$log(L) =$$\lim_{n\to\infty}$$\frac{1}{2n}=0$

Hence, $L=e^0=1$