**Problem 26.** Show that the series is divergent using the Integral Test:\[\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}}\]**Explanation:**To determine if the series diverges or converges, apply the Integral Test. The Integral Test states that for a function $ f(n) = \frac{1}{\sqrt{n}} $, which is positive, continuous, and decreasing for $ n \geq 2 $, we can consider the following improper integral for $ x \geq 2 $:\[\int_{2}^{\infty} \frac{1}{\sqrt{x}} \, dx\]Calculate the integral:1. Use the substitution $ u = \sqrt{x} $, so $ x = u^2 $ and $ dx = 2u \, du $.2. Substitute and integrate:\[\int \frac{1}{\sqrt{x}} \, dx = \int \frac{1}{u} \cdot 2u \, du = 2\int 1 \, du = 2u = 2\sqrt{x}\]3. Evaluate the bounds from 2 to $\infty$:\[\left[ 2\sqrt{x} \right]_{2}^{\infty} = \lim_{b \to \infty} (2\sqrt{b} - 2\sqrt{2})\]The limit $\lim_{b \to \infty} 2\sqrt{b} = \infty$.Since this integral diverges, by the Integral Test, the series $\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}}$ is also divergent.

Answered: 26. Show that the series is divergent…

Math

Advanced Math

26. Show that the series is divergent using the Integral test 1 n=2

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

**Problem 26.** Show that the series is divergent using the Integral Test:

\[
\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}}
\]

**Explanation:**

To determine if the series diverges or converges, apply the Integral Test. The Integral Test states that for a function $ f(n) = \frac{1}{\sqrt{n}} $, which is positive, continuous, and decreasing for $ n \geq 2 $, we can consider the following improper integral for $ x \geq 2 $:

\[
\int_{2}^{\infty} \frac{1}{\sqrt{x}} \, dx
\]

Calculate the integral:

1. Use the substitution $ u = \sqrt{x} $, so $ x = u^2 $ and $ dx = 2u \, du $.
2. Substitute and integrate:

\[
\int \frac{1}{\sqrt{x}} \, dx = \int \frac{1}{u} \cdot 2u \, du = 2\int 1 \, du = 2u = 2\sqrt{x}
\]

3. Evaluate the bounds from 2 to $\infty$:

\[
\left[ 2\sqrt{x} \right]_{2}^{\infty} = \lim_{b \to \infty} (2\sqrt{b} - 2\sqrt{2})
\]

The limit $\lim_{b \to \infty} 2\sqrt{b} = \infty$.

Since this integral diverges, by the Integral Test, the series $\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}}$ is also divergent.

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