Show that $(\sqrt{n + 1} - \sqrt{n})$ converges to 0.

For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template and an example, see this.

Problem Discussion

We’ll follow a similar approach to proving the example in here. We want to show that $(\sqrt{n + 1} - \sqrt{n})$ converges to 0, so we need to find $N \in \mathbf{N}$ such that for any $\epsilon > 0$,

$$ \begin{align*} \lvert \sqrt{n + 1} - \sqrt{n} - 0 \rvert < \epsilon \quad \text{whenever $n > N$}. \end{align*} $$

since $\sqrt{n + 1} > \sqrt{n}$, we can drop the absolute value and then we’ll use the trick of multiplying by the conjugate.

$$ \begin{align*} \lvert \sqrt{n + 1} - \sqrt{n} \rvert &< \epsilon \\ \left(\frac{\sqrt{n + 1} + \sqrt{n}}{\sqrt{n + 1} + \sqrt{n}}\right) \sqrt{n + 1} - \sqrt{n} &< \epsilon \\ \frac{1}{\sqrt{n + 1} + \sqrt{n}} &< \epsilon. \end{align*} $$

We need to remember here that we’re trying to find an $n$ that would make the above inequality true. So we want to solve for $n$. but it’s not easy since we have $\sqrt{n + 1} + \sqrt{n}$ in the denominator. What can we do to simplify this? The goal here is that want to keep this term under $\epsilon$ so instead of focusing on the term itself, we could maybe find a slightly bigger term that is easier to work with. For example we can simplify the bound to:

$$ \begin{align*} \frac{1}{\sqrt{n + 1} + \sqrt{n}} &< \frac{1}{\sqrt{n} + \sqrt{n}} \\ &< \frac{1}{2\sqrt{n}}. \end{align*} $$

$(\frac{1}{2\sqrt{n}})$ is slightly bigger than the original term and so we’re getting an even bigger term to be less than $\epsilon$ so this definitely works.

$$ \begin{align*} \frac{1}{2\sqrt{n}} &< \epsilon \\ \frac{1}{2\epsilon} &< \sqrt{n} \\ \sqrt{n} &> \frac{1}{2\epsilon} \\ n &> \frac{1}{4\epsilon^2}. \end{align*} $$

So now if we let $n$ be greater than $N = \frac{1}{4\epsilon^2}$, we’ll get the bound that we want $\lvert \sqrt{n + 1} - \sqrt{n} \rvert < \epsilon$. In the formal proof, we’ll have to walk backwards.

Formal Proof

To show that $(\sqrt{n + 1} - \sqrt{n})$ converges to 0, let $\epsilon > 0$ be arbitrary. We want to prove that there exists some $N \in \mathbf{N}$ such that when $n > N$,

$$ \begin{align*} \lvert \sqrt{n + 1} - \sqrt{n} - 0\rvert < \epsilon. \end{align*} $$

Pick $N = \frac{1}{4\epsilon^2}$. We now verify that this choice is appropriate. Let $n > N$. Then,

$$ \begin{align*} \lvert \sqrt{n + 1} - \sqrt{n} \rvert \\ \left(\frac{\sqrt{n + 1} + \sqrt{n}}{\sqrt{n + 1} + \sqrt{n}}\right) \sqrt{n + 1} - \sqrt{n} \\ \frac{1}{\sqrt{n + 1} + \sqrt{n}} &< \frac{1}{2\sqrt{n}}. \end{align*} $$

But we know since $n > N$ then $\frac{1}{2\sqrt{n}} < \frac{1}{2\sqrt{N}}$ and so

$$ \begin{align*} \frac{1}{2\sqrt{n}} < \frac{1}{2\sqrt{N}} &< \frac{1}{2\sqrt{\frac{1}{4\epsilon^2}}} \\ &= \frac{1}{\frac{2}{2\epsilon}} \\ &= \epsilon. \\ \end{align*} $$

and we $\lvert \sqrt{n + 1} - \sqrt{n} \rvert \leq \epsilon$ as desired. $\blacksquare$

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