Chapter 8: Q 44. (page 670)

Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$

Short Answer

Expert verified

The interval of convergence for power series is $R$ .

Step by step solution

Step 1. Given information.

The given power series is $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$ .

Step 3. Find the interval of convergence.

Let us assume $b_{k} = \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$ and $b_{k + 1} = \frac{{(- 1)}^{k + 1}}{[2 (k + 1) + 1]!} {(3 x + 7)}^{2 (k + 1) + 1}$

Ratio for the absolute convergence is

role="math" localid="1649425467628" $\lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k \to \infty} |\frac{\frac{{(- 1)}^{k + 1}}{[2 (k + 1) + 1]!} {(3 x + 7)}^{2 k + 3}}{\frac{{(- 1)}^{k}}{[2 k + 1]!} {(3 x + 7)}^{2 k + 1}}| = \lim_{k \to \infty} {|3 x + 7|}^{2} \frac{- 1}{(2 k + 3) (2 k + 2)}$

Now, we evaluate the limit at $k \to \infty$ .

So, $\lim_{k \to \infty} {|3 x + 7|}^{2} \frac{- 1}{(2 k + 3) (2 k + 2)} = 0$ , that is the value of limit will be zero no matter what value the variable $x$ takes.

Hence by the ratio test the series converges absolutely for every value of $x$ .

Therefore, the interval of convergence of the power series is $R$ .

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Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$

Short Answer

Step by step solution

Step 1. Given information.

Step 3. Find the interval of convergence.

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Step by step solution

Step 1. Given information.

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Most popular questions from this chapter

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Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$