Chapter 1: Q24P (page 1)

Use the ratio test to find whether the following series converge or diverge:
24. $\sum_{n = 0}^{\infty} \frac{3^{2 n}}{2^{3 n}}$

Short Answer

Expert verified

The series $\sum_{n = 0}^{\infty} \frac{3^{2 n}}{2^{3 n}}$ divergence.

Step by step solution

Process of ratio test

Apply ratio test in the given series by using $ρ_{n} = | \frac{a_{n + 1}}{a_{n}} |$ and $ρ = \underset{n \to \infty}{l i m} ρ_{n}$ , where $a_{n + 1}$ is the ${(n + 1)}^{t h}$ term of the series and $a_{n}$ is the $n^{t h}$ term. If $ρ < 1$ , then the series converges. If $ρ > 1$ , then the series diverges.

Apply the ratio test

The given series is $\sum_{n = 0}^{\infty} \frac{3^{2 n}}{2^{3 n}}$ .

So, $a_{n + 1} = \frac{3^{2 (n + 1)}}{2^{3 (n + 1)}}$ , and role="math" $a_{n} = \frac{3^{2 n}}{2^{3 n}}$ .

Obtain the value of $ρ_{n} = |\frac{a_{n + 1}}{a_{n}}|$ .

$\begin{array}{rcl} ρ_{n} & = & |\frac{a_{n + 1}}{a_{n}}| \\ = & |\frac{3^{2 (n + 1)}}{2^{3 (n + 1)}} \div \frac{3^{2 n}}{2^{3 n}}| \\ = & \frac{3^{2 (n + 1)}}{2^{3 (n + 1)}} \times \frac{2^{3 n}}{3^{2 n}} \\ = & \frac{3^{2}}{2^{3}} \\ = & \frac{9}{8} \end{array}$

Solve the limit

Now, $ρ = \lim_{n \to \infty} ρ_{n}$ is calculated as follows:

$\begin{array}{rcl} ρ & = & \lim_{n \to \infty} \frac{9}{8} \\ = & \frac{9}{8} \end{array}$

Here, $ρ > 1$ , therefore the series diverges.

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Use the ratio test to find whether the following series converge or diverge:
24. $\sum_{n = 0}^{\infty} \frac{3^{2 n}}{2^{3 n}}$

Short Answer

Step by step solution

Process of ratio test

Apply the ratio test

Solve the limit

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Use the ratio test to find whether the following series converge or diverge:24.∑n=0∞32n23n

Short Answer

Step by step solution

Process of ratio test

Apply the ratio test

Solve the limit

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Work Energy and Power

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Study anywhere. Anytime. Across all devices.

Use the ratio test to find whether the following series converge or diverge:
24. $\sum_{n = 0}^{\infty} \frac{3^{2 n}}{2^{3 n}}$