Find the radius of convergence, R, of the following series. 00 2 n!(9x - 1)" R =

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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**Problem Statement:**

Find the radius of convergence, \( R \), of the following series:

\[
\sum_{n=1}^{\infty} n!(9x - 1)^n
\]

\( R = \) [Enter your answer here]

---

**Task:**

Find the interval, \( I \), of convergence of the series.

**Options:**

- \(\bigcirc\) \( I = \left( -\frac{1}{9}, 0 \right) \cup \left\{ \frac{1}{9} \right\}\)
- \(\bigcirc\) \( I = \{ 1 \} \)
- \(\bigcirc\) \( I = \left\{ \frac{1}{9} \right\} \)
- \(\bigcirc\) \( I = \{ 0 \} \)
- \(\bigcirc\) \( I = \left[ -\frac{1}{9}, \frac{1}{9} \right] \)
Transcribed Image Text:**Problem Statement:** Find the radius of convergence, \( R \), of the following series: \[ \sum_{n=1}^{\infty} n!(9x - 1)^n \] \( R = \) [Enter your answer here] --- **Task:** Find the interval, \( I \), of convergence of the series. **Options:** - \(\bigcirc\) \( I = \left( -\frac{1}{9}, 0 \right) \cup \left\{ \frac{1}{9} \right\}\) - \(\bigcirc\) \( I = \{ 1 \} \) - \(\bigcirc\) \( I = \left\{ \frac{1}{9} \right\} \) - \(\bigcirc\) \( I = \{ 0 \} \) - \(\bigcirc\) \( I = \left[ -\frac{1}{9}, \frac{1}{9} \right] \)
**Problem Statement:**

Find the radius of convergence, \( R \), of the series.

\[
\sum_{n=1}^{\infty} \frac{(x - 5)^n}{n^n}
\]

**Task 1: Radius of Convergence**

\( R = \underline{\hspace{2cm}} \)

**Task 2: Interval of Convergence**

Find the interval, \( I \), of convergence of the series. (Enter your answer using interval notation.)

\( I = \underline{\hspace{2cm}} \)
Transcribed Image Text:**Problem Statement:** Find the radius of convergence, \( R \), of the series. \[ \sum_{n=1}^{\infty} \frac{(x - 5)^n}{n^n} \] **Task 1: Radius of Convergence** \( R = \underline{\hspace{2cm}} \) **Task 2: Interval of Convergence** Find the interval, \( I \), of convergence of the series. (Enter your answer using interval notation.) \( I = \underline{\hspace{2cm}} \)
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