Difference Between Adjacent Square Roots Converges
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Theorem
Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as $x_n = \sqrt {n + 1} - \sqrt n$.
Then $\left \langle {x_n} \right \rangle$ converges to a zero limit.
Proof
We have:
| \(\displaystyle \) | \(\displaystyle 0\) | \(\le\) | \(\displaystyle \sqrt {n + 1} - \sqrt n\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left({\sqrt {n + 1} - \sqrt n}\right) \left({\sqrt {n + 1} + \sqrt n}\right)} {\sqrt {n + 1} + \sqrt n}\) | \(\displaystyle \) | multiplying top and bottom by $\sqrt {n + 1} - \sqrt n$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {n + 1 - n} {\sqrt {n + 1} + \sqrt n}\) | \(\displaystyle \) | Difference of Two Squares | ||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {\sqrt {n + 1} + \sqrt n}\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(<\) | \(\displaystyle \frac 1 {\sqrt n}\) | \(\displaystyle \) | as $\sqrt {n + 1} + \sqrt n > \sqrt n$ |
But from Power of Reciprocal, $\dfrac 1 {\sqrt n} \to 0$ as $n \to \infty$.
The result follows by the Squeeze Theorem.
$\blacksquare$
Sources
- K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 4.20 \ (3)$
