Chapter 8: Q 46. (page 670)

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

Short Answer

Expert verified

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

Step by step solution

Step 2. Find the interval of convergence.

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

Ratio for the absolute convergence is

$\lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}|$ =

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

Since for $k \to \infty$ the limit is zero irrespective of the value of variable.

This implies that

$\lim_{k \to \infty} |\frac{1}{2 k + 3} x| = 0$

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

Short Answer

Step by step solution

Step 2. Find the interval of convergence.

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