Question

For which real numbers x does the series ∑_(n=1) to n = infinity (√(n+1)-√n)/n^x converges?

Answer 1

Answer on question 40091 – Math – Real Analysis

For which real numbers $x$ does the series

$$\sum_{n=1}^{\infty} \frac{\sqrt{n+1} - \sqrt{n}}{n^x}$$

converges?

Solution:

Consider the following series

$$\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}.$$

This series converges if $\alpha > 1$. The series

$$\sum_{n=1}^{\infty} \frac{\sqrt{n+1} - \sqrt{n}}{n^x}$$

converges if

$$\lim_{n \to \infty} \frac{\sqrt{n+1} - \sqrt{n}}{n^x} : \frac{1}{n^{\alpha}} < \infty, \quad \alpha > 1.$$

Thus we have

$$\begin{aligned} \lim_{n \to \infty} \frac{\sqrt{n+1} - \sqrt{n}}{n^x} : \frac{1}{n^{\alpha}} &= \lim_{n \to \infty} \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}} \cdot n^{\alpha - x} = \\ &= \lim_{n \to \infty} \frac{n+1 - n}{\sqrt{n} \left( \sqrt{\frac{n+1}{n}} + 1 \right)} \cdot n^{\alpha - x} = \lim_{n \to \infty} \frac{1}{n^{0.5} \left( \sqrt{1 + \frac{1}{n}} + 1 \right)} \cdot n^{\alpha - x} = \\ &= \frac{1}{\sqrt{1 + 0} + 1} \lim_{n \to \infty} n^{\alpha - x - 0.5} = \frac{1}{2} \lim_{n \to \infty} n^{\alpha - x - 0.5}. \end{aligned}$$

We know that $\lim_{n \to \infty} n^{\alpha - x - 0.5} < \infty$ if $(\alpha - x - 0.5) \leq 0$. Thus we have next system of conditions

$$\left\{ \begin{array}{c} \alpha - x - 0.5 \leq 0, \\ \alpha > 1, \end{array} \right. \Rightarrow \left\{ \begin{array}{c} \alpha - 0.5 \leq x, \\ \alpha > 1, \end{array} \right. \Rightarrow x \geq \alpha - 0.5 > 1 - 0.5 = 0.5$$

Thus we have that

$$\boxed{x > 0.5}$$

Answer:

$$\boxed{x > 0.5}$$