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.

$\blacksquare$


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