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.