Chapter 8: Q 48. (page 670)

Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k}$

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 $\sum_{k = 0}^{\infty} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k}$ .

Step 2. Find the interval of convergence.

Let us assume $b_{k} = \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k}$ therefore $b_{k + 1} = \frac{{(k + 1)}^{3}}{1.3.5 . . . . . (2 (k + 1) + 1)} {(x + 1)}^{k + 1}$

Ratio for the absolute convergence is

\lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k \to \infty} |\frac{\frac{{(k + 1)}^{3}}{1.3.5 . . . . (2 (k + 1) + 1)} {(x + 1)}^{k + 1}}{\frac{k^{3}}{1.3.5 . . . . . (2 k + 1)} {(x + 1)}^{k}}| = \lim_{k \to \infty} |\frac{1}{2 k + 3} {(\frac{k}{k + 3})}^{3} (x + 1)|

So, by ratio test of absolute convergence the series will converge when $|x + 1| < 1$ .

This implies that

$- 1 < x + 1 < 1$

So, $- 1 < x + 1$ and $x + 1 < 1$

$x > - 2$ and $x < 0$

Step 2. Find the interval of convergence.

Evaluate the series at $x = - 2$

$\sum_{k = 0}^{\infty} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k} = \sum_{k = 0}^{\infty} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(- 2 + 1)}^{k} {= \sum}_{k = 0}^{\infty} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(- 1)}^{k}$

Thus, the series will diverge.

Evaluate the series when $x = 0$ .

$\sum_{k = 0}^{\infty} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k} = \sum_{k = 0}^{\infty} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(0 + 1)}^{k} = \sum_{k = 0}^{\infty} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(1)}^{k}$

Thus, the series will diverge.

Therefore the value of for which the series converges is $(- 2, 0)$ .

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Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Find the interval of convergence.

Step 2. Find the interval of convergence.

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Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k}$