Find the radius of convergence, $R$, of the series: $\displaystyle\sum_{n=1}^{\infty} \frac{8^n (x+8)^n}{\sqrt{n}}$

$R= \boxed{}$

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

$I= \boxed{}$

Neetesh Kumar | December 7, 2024

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This is the solution to Math 132
Assignment: 11.8 Question Number 8
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Step-by-step solution:

To find the radius of convergence $R$, we apply the Ratio Test to the general term:

$a_n = \frac{8^n (x+8)^n}{\sqrt{n}}.$

Step 1: Apply the Ratio Test

The Ratio Test states that the series converges absolutely if:

$\displaystyle\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1.$

Compute the ratio:

$\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{8^{n+1} (x+8)^{n+1}}{\sqrt{n+1}}}{\frac{8^n (x+8)^n}{\sqrt{n}}} \right| = \left| 8 \cdot (x+8) \cdot \frac{\sqrt{n}}{\sqrt{n+1}} \right|.$

Simplify the terms:

$\left| \frac{a_{n+1}}{a_n} \right| = \left| 8(x+8) \right| \cdot \frac{\sqrt{n}}{\sqrt{n+1}}.$

Step 2: Take the limit as $n \to \infty$

For large $n$, the term $\frac{\sqrt{n}}{\sqrt{n+1}}$ approaches $1$.

Thus:

$\displaystyle\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| 8(x+8) \right|.$

The Ratio Test guarantees convergence if:

$\left| 8(x+8) \right| < 1.$

Simplify the inequality:

$\left| x+8 \right| < \frac{1}{8}.$

Thus, the radius of convergence is:

$R = \boxed{\frac{1}{8}}$

Step 3: Interval of convergence

The series converges absolutely for $\left| x+8 \right| < \frac{1}{8}$, which gives the interval:

$-8 - \frac{1}{8} < x < -8 + \frac{1}{8},$ or equivalently:

$-\frac{65}{8} < x < -\frac{63}{8}.$

Next, we test the endpoints $x = -\frac{65}{8}$ and $x = -\frac{63}{8}$ to determine whether the series converges there.

Case 1: $x = -\frac{65}{8}$

Substitute $x = -\frac{65}{8}$ into the series:

$\displaystyle\sum_{n=1}^\infty \frac{8^n (-\frac{65}{8}+8)^n}{\sqrt{n}} = \displaystyle\sum_{n=1}^\infty \frac{8^n (-\frac{1}{8})^n}{\sqrt{n}} = \displaystyle\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}.$

This is an alternating series with terms $a_n = \frac{1}{\sqrt{n}}$.

Since the terms decrease monotonically to $0$, the series converges by the Alternating Series Test.

Case 2: $x = -\frac{63}{8}$

Substitute $x = -\frac{63}{8}$ into the series:

$\displaystyle\sum_{n=1}^\infty \frac{8^n (-\frac{63}{8}+8)^n}{\sqrt{n}} = \displaystyle\sum_{n=1}^\infty \frac{8^n (\frac{1}{8})^n}{\sqrt{n}} = \displaystyle\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.$

This is the harmonic series with $p = \frac{1}{2}$, which diverges because $p \leq 1$.

Final Interval of Convergence

The series converges for $x \in \left[-\frac{65}{8}, -\frac{63}{8}\right)$.

$I = \boxed{\left[-\frac{65}{8}, -\frac{63}{8}\right)}$

Final Answers:

Radius of convergence:

$R = \boxed{\frac{1}{8}}$

Interval of convergence:

$I = \boxed{\left[-\frac{65}{8}, -\frac{63}{8}\right)}$

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