Chapter 2: Q10P (page 59)

Find the disk of convergence for each of the following complex power series.
$\sum_{n = 1}^{\infty} \frac{{(i z)}^{n}}{n^{2}}$

Short Answer

Expert verified

The required disk of convergence is . $| z | < 1$

Step by step solution

Determine Disk of Convergence

For any power series $\sum a_{n} z^{n}$ where z is a complex numbers, then disk of convergence is given by:. $ρ = \underset{n \to \infty}{l i m} | \frac{z \times n}{n + 1} | = | z |$

Step 2:Find the disk of Convergence

The given power series is:, $\sum_{n = 1}^{\infty} \frac{{(i z)}^{n}}{n^{2}}$ where, . $a_{n} = \frac{{(i z)}^{n}}{n^{2}}$

Now, let us evaluate the ratio as:

$\begin{matrix} ρ = \lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | \\ = \lim_{n \to \infty} | \frac{\frac{{(i z)}^{n + 1}}{{(n + 1)}^{2}}}{\frac{{(i z)}^{n}}{n^{2}}} | \\ = \lim_{n \to \infty} | i z {(\frac{n}{n + 1})}^{2} | \end{matrix}$

Now, for the series to be convergent, we have . $ρ < 1$ So,

$\begin{matrix} ρ = \lim_{n \to \infty} | i z {(\frac{n}{n + 1})}^{2} | < 1 \\ | i z | < \lim_{n \to \infty} | {(\frac{n}{n + 1})}^{2} | \\ | z | < \lim_{n \to \infty} | {(\frac{n}{n + 1})}^{2} | \\ | z | < 1 \end{matrix}$

Hence, the required disk of convergence is . $| z | < 1$

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Find the disk of convergence for each of the following complex power series.
$\sum_{n = 1}^{\infty} \frac{{(i z)}^{n}}{n^{2}}$

Short Answer

Step by step solution

Determine Disk of Convergence

Step 2:Find the disk of Convergence

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Find the disk of convergence for each of the following complex power series.∑n=1∞(iz)nn2

Short Answer

Step by step solution

Determine Disk of Convergence

Step 2:Find the disk of Convergence

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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Study anywhere. Anytime. Across all devices.

Find the disk of convergence for each of the following complex power series.
$\sum_{n = 1}^{\infty} \frac{{(i z)}^{n}}{n^{2}}$