limit of nth root of n

The $n$ th root (http://planetmath.org/NthRoot) 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 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.