Radius and Interval of Convergence of Power Series

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AP Calculus BC › Radius and Interval of Convergence of Power Series

Questions 1 - 3

Which of following intervals of convergence cannot exist?

$\small [2,\infty)$

$\small [2,10000]$

For any $\small q,p$ such that $\small q \leq p$ , the interval

For any $\small \epsilon>0$ , the interval $\small [a-\epsilon,a+\epsilon]$ for some $\small a\in\mathbb{R}$

Explanation

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

$\left ( -\infty , \infty \right )$

$\left ( \frac{-1}{3}, \frac{1}{3} \right )$

$\left [ \frac{-1}{3}, \frac{1}{3} \right ]$

$\left ( -3, 3\right )$

Explanation

Using the root test,

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0$

Because 0 is always less than 1, the root test shows that the series converges for any value of x.

Therefore, the interval of convergence is:

$(-\infty, \infty)$

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

$\left ( -5, 5\right )$

$\left ( \frac{-1}{5}, \frac{1}{5} \right )$

$(-\infty, \infty)$

Explanation

Using the root test

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$

and

$\lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$ . T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

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