Determine the convergence or divergence of f(n) = 2^1/n

by RoRi

February 27, 2016

Consider the function $f(n)$ defined by

$f(n) = 2^{\frac{1}{n}}.$

Determine whether the sequence $\{ f(n) \}$ converges or diverges, and if it converges find the limit.

We know that the $n$ th root function is strictly increasing for positive real values (page 147 of Apostol). Therefore,

$1^{\frac{1}{n}} \leq 2^{\frac{1}{n}} \leq n^{\frac{1}{n}} \qquad \text{for all } n \geq 2.$

From property (10.12) (on page 380 of Apostol) we know

$\lim_{n \to \infty} n^{\frac{1}{n}} = 1.$

Hence, by the squeeze theorem (and since $\lim_{n \to \infty} 1^{\frac{1}{n}} = 1$ , we have

$\lim_{n \to \infty} 2^{\frac{1}{n}} = 1.$

Therefore the sequence $\{ f_n \}$ converges with limit 1.

Stumbling Robot