Advadlippabrj

Answered

2022-01-24

Why does
$\sqrt{n\sqrt{n\sqrt{n\dots }}}=n$

Answer & Explanation

Jason Duke

Expert

2022-01-25Added 11 answers

$X=\sqrt{n\cdot \sqrt{n\cdot \sqrt{n\dots }}}$
$=\sqrt{n}\cdot \sqrt{\sqrt{n}}\cdot \sqrt{\sqrt{\sqrt{n}}}\cdot \dots$
$={n}^{\frac{1}{2}}\cdot {n}^{\frac{1}{4}}\cdot {n}^{\frac{1}{8}}\dots$
$={n}^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\dots }$
$={n}^{1}$
$=n$

SaphypycleDapc5

Expert

2022-01-26Added 11 answers

Suppose $y=\sqrt{x\sqrt{x\sqrt{x}}}$
Multiply both sides by x and take the square root:
$\sqrt{xy}=\sqrt{x\sqrt{x\sqrt{x...}}}=y$
Therefore, $\sqrt{xy}=y$, and solving we have $xy={y}^{2}⇒x=y$

RizerMix

Expert

2022-01-27Added 437 answers

It is important to show that the limit exists. Let define the sequence ${a}_{k}=\sqrt{n{a}_{k-1}}$ Since $\frac{{a}_{k}}{{a}_{k-1}}=\sqrt{\frac{n}{{a}_{k-1}}}$ and $\frac{{a}_{k}}{n}=\sqrt{\frac{{a}_{k-1}}{n}}$, we have 1. if ${a}_{k-1}\le n$, then ${a}_{k-1}\le {a}_{k}\le n$; that is, ${a}_{k}$ is increasing and bounded above by n. 2. if ${a}_{k-1}\ge n$, then ${a}_{k-1}\ge {a}_{k}\ge n$; that is, ${a}_{k}$ is decreasing and bounded below by n. In either case, ${a}_{k}$ is convergent. Using the continuity of multiplication by a constant and the continuity of square root, we get $\underset{k\to \mathrm{\infty }}{lim}{a}_{k}=\underset{k\to \mathrm{\infty }}{lim}\sqrt{n{a}_{k-1}}=\sqrt{n\underset{k\to \mathrm{\infty }}{lim}{a}_{k}}$ Squaring and dividing by $\underset{k\to \mathrm{\infty }}{lim}{a}_{k}$, we get that $\underset{k\to \mathrm{\infty }}{lim}{a}_{k}=n$

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