# Difference Between Adjacent Square Roots Converges

## 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.

## Proof

We have:

 $\displaystyle 0$ $\le$ $\displaystyle \sqrt {n + 1} - \sqrt n$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {\sqrt {n + 1} - \sqrt n} \paren {\sqrt {n + 1} + \sqrt n} } {\sqrt {n + 1} + \sqrt n}$ multiplying top and bottom by $\sqrt {n + 1} - \sqrt n$ $\displaystyle$ $=$ $\displaystyle \frac {n + 1 - n} {\sqrt {n + 1} + \sqrt n}$ Difference of Two Squares $\displaystyle$ $=$ $\displaystyle \frac 1 {\sqrt {n + 1} + \sqrt n}$ $\displaystyle$ $<$ $\displaystyle \frac 1 {\sqrt n}$ 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.

$\blacksquare$