Determine the convergence or divergence of f(n) = (n+1)1/2 - n1/2

Consider the function $f(n)$ defined by

$f(n) = \sqrt{n+1} - \sqrt{n}.$

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

First, we use multiply the numerator and denominator by $\sqrt{n+1} + \sqrt{n}$ ,

$\begin{align*} f(n) &= \sqrt{n+1} - \sqrt{n} \\[9pt] &= \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}} \\[9pt] &= \frac{n+1 - n}{\sqrt{n+1} + \sqrt{n}} \\[9pt] &= \frac{1}{\sqrt{n+1} + \sqrt{n}}. \end{align*}$

Then, since $\sqrt{n+1} > \sqrt{n}$ we have

$0 < \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2 \sqrt{n}}.$

Since we know from property (10.9) on page 380 of Apostol that

$\lim_{n \to \infty} \frac{1}{n^{\alpha}} = 0 \qquad \text{if } \alpha > 0$

we then have

$\lim_{n \to \infty} \frac{1}{2 \sqrt{n}} = 0.$

Hence, by the squeeze theorem we have

$\lim_{n \to \infty} f(n) = 0.$

Therefore the sequence $\{ f(n) \}$ converges with limit 0.

Stumbling Robot

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

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