26. Show that the series is divergent using the Integral test 1 n=2
Advanced Engineering Mathematics
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Chapter2: Second-order Linear Odes
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![**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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2396c550-bba7-4086-87a5-883b238a150c%2Ff7896232-80a6-4735-bd3c-7e6a41f66b7e%2F5sls8tv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**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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