Question

Q. Let yn=√(n+1)-√n for nϵN. Show that (yn) and (√nyn) converge. Find their limits.

Answer 1

Answer on Question #64529 – Math – Real Analysis

Question

Let $y_n = \sqrt{n + 1} - \sqrt{n}$ for $n \in \mathbb{N}$. Show that $y_n$ and $\sqrt{n} \cdot y_n$ converge. Find their limits.

Solution

$$y_n = \sqrt{n + 1} - \sqrt{n} = \frac{(\sqrt{n + 1} - \sqrt{n})(\sqrt{n + 1} + \sqrt{n})}{\sqrt{n + 1} + \sqrt{n}} = \frac{n + 1 - n}{\sqrt{n + 1} + \sqrt{n}} = \frac{1}{\sqrt{n + 1} + \sqrt{n}} < \frac{1}{\sqrt{n}}, \quad n \in \mathbb{N}.$$

Obviously $\frac{1}{\sqrt{n}} \to 0$ as $n \to \infty$ because for all $\varepsilon > 0 \ni N = N(\varepsilon)$: $\forall n > N(\varepsilon) \frac{1}{\sqrt{n}} < \varepsilon$ (for example $N(\varepsilon) = \left[\frac{1}{\varepsilon^2}\right] + 1$ where $\left[\frac{1}{\varepsilon^2}\right]$ denotes the integer part of $\frac{1}{\varepsilon^2}$).

So we have

$$0 < \frac{1}{\sqrt{n + 1} + \sqrt{n}} < \frac{1}{\sqrt{n}}.$$

Using squeeze theorem we obtain that

$$\frac{1}{\sqrt{n + 1} + \sqrt{n}} \to 0 \quad \text{as} \quad n \to \infty.$$

Thus $\lim_{n \to \infty} y_n = \lim_{n \to \infty} \left( \sqrt{n + 1} - \sqrt{n} \right) = 0$.

Next,

$$\sqrt{n} \cdot y_n = \frac{\sqrt{n}}{\sqrt{n + 1} + \sqrt{n}} = \frac{\sqrt{n}}{\sqrt{n \left(1 + \frac{1}{n}\right) + \sqrt{n}}} = \frac{\sqrt{n}}{\sqrt{n} \left( \sqrt{1 + \frac{1}{n} + 1} \right)} = \frac{1}{\sqrt{1 + \frac{1}{n} + 1}}.$$

Obviously $\frac{1}{n} \to 0$ as $n \to \infty$ because for all $\varepsilon > 0 \ni N = N(\varepsilon)$: $\forall n > N(\varepsilon) \frac{1}{n} < \varepsilon$ (for example, $N(\varepsilon) = \left[\frac{1}{\varepsilon}\right] + 1$, where $\left[\frac{1}{\varepsilon}\right]$ denotes the integer part of $\frac{1}{\varepsilon}$).

Then

$$\frac{1}{\sqrt{1 + \frac{1}{n} + 1}} \to \frac{1}{\sqrt{1 + 1}} = \frac{1}{2} \quad \text{as} \quad n \to \infty.$$

Thus $\lim_{n \to \infty} \sqrt{n} \cdot y_n = \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n + 1} + \sqrt{n}} = \frac{1}{2}$.

**Answer**: $\lim_{n \to \infty} y_n = 0$; $\lim_{n \to \infty} \sqrt{n} \cdot y_n = \frac{1}{2}$.

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