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Determine the convergence or divergence of f(n) = 21/n

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 nth 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.

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