Chapter 1: Q19P (page 1)

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

Short Answer

Expert verified

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

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 $ρ = \lim_{n \to \infty} ρ_{n}$ , where $a_{n + 1}$ is the ${(n + 1)}^{th}$ term of the series and $a_{n}$ is the $n^{th}$ term. If role="math" localid="1664190384146" $ρ<1$ , then the series converges. If role="math" $ρ>1$ , then the series diverges.

Apply the ratio test

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

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

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

$\begin{array}{rcl} ρ_{n} & = & |\frac{a_{n + 1}}{a_{n}}| \\ = & |\frac{3^{n + 1}}{2^{2 (n + 1)}} \div \frac{3^{n}}{2^{2 n}}| \\ = & \frac{3^{n + 1}}{2^{2 (n + 1)}} \times \frac{2^{2 n}}{3^{n}} \\ = & \frac{3}{4} \end{array}$

Solve the limit

Now, is calculated as follows:

$ρ= \lim_{n \to \infty} ρ_{n} = \lim_{n \to \infty} \frac{3}{4} = \frac{3}{4}$

Here, $ρ < 1$ , therefore the series converges.

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

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:19.∑n=0∞3n22n

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

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