# Find the interval & radius of convergence for the power series in 1b?

May 23, 2018

interval: $1 \le x < 9$
radius: $4$

#### Explanation:

use the ratio test:

[note: for the series ${\sum}_{n = 1}^{\infty} {a}_{n}$, if ${\lim}_{n \rightarrow \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid$
-is $< 1$, series converges
-is $= 1$, inconclusive
-is $> 1$, series diverges]

if you apply the ratio test to this: ${\sum}_{n = 1}^{\infty} {\left(x - 5\right)}^{n} / \left(n {4}^{n}\right)$, evaluating lim_(nrarroo)abs(((x-5)^(n+1)/((n+1)(4^(n+1))))/((x-5)^(n)/((n)(4^(n)))) will show what x-values make the series converge or diverge.

simplifying the limit: =lim_(nrarroo)abs(((x-5)/((n+1)(4)))/(1/((n)))#
$= {\lim}_{n \rightarrow \infty} \left\mid \frac{n \left(x - 5\right)}{4 n + 4} \right\mid$

[note: here you can divide both the numerator and denominator by $n$, because for whatever $n$ value is used, the value inside the absolute value signs will stay the same (dividing by $\frac{n}{n}$ or 1)]

$= {\lim}_{n \rightarrow \infty} \left\mid \frac{\frac{n \left(x - 5\right)}{n}}{\frac{4 n}{n} + \frac{4}{n}} \right\mid$
$= \left\mid \frac{x - 5}{4 + 0} \right\mid$
$= \left\mid \frac{x - 5}{4} \right\mid$

back to the ratio test, the series can only converge if $\left\mid \frac{x - 5}{4} \right\mid < 1$ or $\left\mid \frac{x - 5}{4} \right\mid = 1$

Case 1: $\left\mid x - 5 \right\mid < 4$
$- 4 < x - 5 < 4$
$1 < x < 9$ (the solution interval must include these values)

Case 2: $\left\mid \frac{x - 5}{4} \right\mid = 1$
$x - 5 = - 4 , x - 5 = 4$
$x = 1 , 9$

if $x = 1$, series becomes: ${\sum}_{n = 1}^{\infty} {\left(1 - 5\right)}^{n} / \left(n {4}^{n}\right)$
$= {\sum}_{n = 1}^{\infty} {\left(- 4\right)}^{n} / \left(n {4}^{n}\right)$
$= {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} / \left(n\right)$
this is the alternating harmonic series, which converges by the alternating series test

if $x = 9$, series becomes: ${\sum}_{n = 1}^{\infty} {\left(9 - 5\right)}^{n} / \left(n {4}^{n}\right)$
$= {\sum}_{n = 1}^{\infty} {\left(4\right)}^{n} / \left(n {4}^{n}\right)$
$= {\sum}_{n = 1}^{\infty} \frac{1}{n}$
this is the harmonic series, which diverges. here is a proof

so include $x = 1$ in the interval, too: $1 \le x < 9$

radius of convergence is half the difference between the upper and lower values for the interval $= \frac{9 - 1}{2} = 4$

and here is a video with a similar problem