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Two part question that I want to make sure I did correctly.

a) Let $x_n = \sqrt[n]{n} - 1$. Use the fact that $(1 + x_n)^n = n$ to show that $x_n^2 \leq \frac{2}{n}$. Hint given to use the binomial theorem and throw away most terms.

Expanding $(1+x_n)^n$ using the binomial theorem gives us ${{n}\choose{0}}x_n^{n-0}1^0 + {{n}\choose{1}}x_n^{n-1}1^1 + \cdots + {{n}\choose{n}}x_n^{0}1^n = n$.

Want to show $x_n^2 \leq 2/n$, that is $(\sqrt[n]{n} - 1)^2 \leq \frac{2}{{{n}\choose{0}}x_n^{n-0}1^0 + {{n}\choose{1}}x_n^{n-1}1^1 + \cdots + {{n}\choose{n}}x_n^{0}1^n}$ or $n^{2/n} - 2\sqrt[n]{n} + 1\leq {{n}\choose{0}}x_n^{n-0}1^0 + {{n}\choose{1}}x_n^{n-1}1^1 + \cdots + {{n}\choose{n}}x_n^{0}1^n$ which we can see is true by getting rid of the right terms

b) Compute $\lim \limits_{n\to \infty} n^{1/n}$

Using our inequality in part a), we can use Squeeze Theorem to compute the limit.

We have $1^{1/n} \leq n^{1/n} \leq 2/n + 1$ Since $1^{1/n} \rightarrow 1$ and $2/n +1 \rightarrow 1$, we have that $ n^{1/n} \rightarrow 1$. So $\lim \limits_{n\to \infty} n^{1/n} = 1$.

Just wanted to double check this.

SS'
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  • It should be: $n^{1/n} \leq \sqrt{2/n} + 1$, since $x_n^2 \leq 2/n$. What is part (a) used for besides part (b)? There are certainly easier ways to prove this such as here. – JavaMan Oct 17 '18 at 02:56
  • Just an exercise in a textbook. Was supposed to prove it like this. But thanks. – SS' Oct 17 '18 at 03:07

2 Answers2

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You do not really need the binomial theorem. For any $n\geq 2$ we have

$$ n = \frac{n}{n-1}\cdot\frac{n-1}{n-2}\cdot\ldots\cdot\frac{2}{1}\cdot \frac{1}{1}\tag{1}$$ hence $n^{1/n}$ is the geometric mean of $1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}$, which by the AM-GM inequality is less than $$ \frac{1}{n}\left[1+\sum_{k=1}^{n-1}\left(1+\frac{1}{k}\right)\right]=1+\frac{H_{n-1}}{n}\leq 1+\frac{1+\int_{1}^{n}\frac{dx}{x}}{n}=1+\frac{1+\log(n)}{n}\tag{2} $$ but clearly greater than one. Squeezing then leads to $\lim_{n\to +\infty} n^{1/n}=1$, as expected from the fact that this limit equals $\exp\lim_{n\to +\infty}\frac{\log n}{n}$.

Jack D'Aurizio
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Here is a slight modification of the approach given by the hints:

Note that $$ \begin{align} \left(1+\sqrt{\frac2n}\right)^n &\ge1+\binom{n}{1}\sqrt{\frac2n}+\binom{n}{2}\frac2n\tag{1a}\\ &=n+\sqrt{2n}\tag{1b}\\[6pt] &\ge n\tag{1c} \end{align} $$ Explanation:
$\text{(1a)}$: truncation of the Binomial Theorem
$\text{(1b)}$: $\binom{n}{1}\sqrt{\frac2n}=\sqrt{2n}$ and $\binom{n}{2}\frac2n=n-1$
$\text{(1c)}$: $\sqrt{2n}\ge0$

Relation to the hint: $(1)$ and the definition of $x_n$ imply $$ (1+x_n)^n=n\le\left(1+\sqrt{\frac2n}\right)^n\tag2 $$ $(2)$ says that $x_n\le\sqrt{\frac2n}$ .

Taking the $n^\text{th}$ root of $(1)$ yields $$ 1\le n^{1/n}\le1+\sqrt{\frac2n}\tag3 $$ and by the Squeeze Theorem, $$ \lim_{n\to\infty}n^{1/n}=1\tag4 $$

robjohn
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