Difference Between Adjacent Square Roots Converges

This article was Featured Proof between 12th July 2012 and 22nd July 2012.

Theorem

Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = \sqrt {n + 1} - \sqrt n$.

Then $\sequence {x_n}$ converges to a zero limit.

We have:

$\ds 0$

$\le$

$\ds \sqrt {n + 1} - \sqrt n$

$\ds $

$=$

$\ds \frac {\paren {\sqrt {n + 1} - \sqrt n} \paren {\sqrt {n + 1} + \sqrt n} } {\sqrt {n + 1} + \sqrt n}$

multiplying top and bottom by $\sqrt {n + 1} - \sqrt n$

$\ds $

$=$

$\ds \frac {n + 1 - n} {\sqrt {n + 1} + \sqrt n}$

$\ds $

$=$

$\ds \frac 1 {\sqrt {n + 1} + \sqrt n}$

$\ds $

$<$

$\ds \frac 1 {\sqrt n}$

as $\sqrt {n + 1} + \sqrt n > \sqrt n$

But from Sequence of Powers of Reciprocals is Null Sequence, $\dfrac 1 {\sqrt n} \to 0$ as $n \to \infty$.

The result follows by the Squeeze Theorem.

$\blacksquare$