X=n⋅n⋅n… =n⋅n⋅n⋅… =n12⋅n14⋅n18… =n12+14+18… =n1 =n

Suppose y=xxx Multiply both sides by x and take the square root: xy=xxx...=y Therefore, xy=y, and solving we have xy=y2⇒x=y

X=n⋅n⋅n… =n⋅n⋅n⋅… =n12⋅n14⋅n18… =n12+14+18… =n1 =n

Suppose y=xxx Multiply both sides by x and take the square root: xy=xxx...=y Therefore, xy=y, and solving we have xy=y2⇒x=y

We have an answer to "Why does nnn…=n"

Advadlippabrj

Answered question

2022-01-24

Why does

\sqrt{n \sqrt{n \sqrt{n \dots}}} = n

Answer & Explanation

Jason Duke

Beginner2022-01-25Added 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

Beginner2022-01-26Added 11 answers

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

RizerMix

Expert2022-01-27Added 656 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} \leq n

, then

a_{k - 1} \leq a_{k} \leq n

; that is,

a_{k}

is increasing and bounded above by n. 2. if

a_{k - 1} \geq n

, then

a_{k - 1} \geq a_{k} \geq 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

lim_{k \to \infty} a_{k} = lim_{k \to \infty} \sqrt{n a_{k - 1}} = \sqrt{n lim_{k \to \infty} a_{k}}

Squaring and dividing by

lim_{k \to \infty} a_{k}

, we get that

lim_{k \to \infty} a_{k} = n

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