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Homelimit of nth root of n

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The $n$th root of $n$ tends to 1 as $n$ tends to infinity, i.e. the real number sequence

$\sqrt[1]{1},\,\sqrt[2]{2},\,\sqrt[3]{3},\,\ldots,\,\sqrt[n]{n},\,\ldots$ |

converges to the limit

$\displaystyle\lim_{{n\to\infty}}\sqrt[n]{n}=1.$ | (1) |

Proof. If we denote $\sqrt[n]{n}:=1+\delta_{n}$, we may write by the binomial theorem that

$n=(1+\delta_{n})^{n}=1+{n\choose 1}\delta_{n}+{n\choose 2}\delta_{n}^{2}+% \ldots+{n\choose n}\delta_{n}^{n}.$ |

This implies, since all terms of the right hand side are positive (when $n>1$), that

$n>{n\choose 2}\delta_{n}^{2}=\frac{n(n\!-\!1)}{2!}\delta_{n}^{2},\qquad\delta_% {n}^{2}<\frac{2}{n-1},\qquad 0<\delta_{n}<\sqrt{\frac{2}{n-1}},$ |

whence $\displaystyle\lim_{{n\to\infty}}\delta_{n}=0$. Accordingly,

$\lim_{{n\to\infty}}\sqrt[n]{n}=\lim_{{n\to\infty}}(1+\delta_{n})=1,$ |

Q.E.D.

Note. (1) follows also from the corollary 3 in the entry growth of exponential function.

Synonym:

sequence of nth roots of n

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## Mathematics Subject Classification

12D99*no label found*30-00

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## Corrections

Correction. by fishgoldstein ✘

## Comments

## A puzzle.

Here is a puzzle:

Let's say someone shows you a rod 99 feet in length.

He tells you that you are allowed to cut it up as you see, and you will be given the product of the lengths of all of the pieces in dollars.

For example, if you were to slice it in two pieces 48.5 feet in length, you would be given 48.5 x 48.5 dollars, (2352.25)

How should you cut up the rod?

The answer: make 33 pieces each 3 feet long.

The reason: the nth root of n reaches it's maximum at 3.

## Re: A puzzle.

It is true that the maximum is reached for all pieces having the same length. Then, the product for n pieces is (S/n)^n (S=99). The maximum is reached at n = S/e = 36.42; the closest integer is 36, not 33.

## Re: A puzzle.

Cut one piece an 1/8'' long and another piece 98' 11 7/8'' long.

Then I would pay a dollar. If that is still too much for you,

make the short piece shorter.