# Difference Between Adjacent Square Roots Converges

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## 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.

$\blacksquare$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (3)$