Ratio Test vs AST

[cherry]

Homework Statement

Find the interval of convergence for the given power series.

Relevant Equations

N/A

Hi, I'm having difficulty understanding why the interval of convergence is (0, 18].
When I tested x=18, I got the following conclusion using the ratio test.

When I attempt using AST, the function still diverges as the lim (n -> inf) = 2^n / n ≠ 0.
What am I missing?

Thanks!

 

[Mark44]

Homework Statement: Find the interval of convergence for the given power series.
Relevant Equations: N/A

Hi, I'm having difficulty understanding why the interval of convergence is (0, 18].
When I tested x=18, I got the following conclusion using the ratio test.
View attachment 344579

When I attempt using AST, the function still diverges as the lim (n -> inf) = 2^n / n ≠ 0.
What am I missing?

Thanks!

The series is this:
$$\sum_{n = 1}^\infty \frac{(x - 9)^n}{n(-9)^n}$$
When you substitute x = 18, the numerator is not $18^n$.

 

[cherry]

The series is this:
$$\sum_{n = 1}^\infty \frac{(x - 9)^n}{n(-9)^n}$$
When you substitute x = 18, the numerator is not $18^n$.

Oh my gosh, thank you!

 

[FactChecker]

One of the most important things to learn from mathematics is to be careful about each step. It is definitely a learning process. Mathematics is one subject where you need to get a long string of steps correct to get the right answer, and it is fairly unique in that respect. Also, when you are thinking about the hard steps, it is often the easy ones where mistakes occur. So you should make it a habit to review your work with as much attention to the easy steps as you give to the hard steps.

 

[nuuskur]

It's a power series at the point $a=9$ whose radius of convergence $R$ is given by
$$\frac{1}{R} = \limsup _n \frac{1}{\sqrt[n]{n9^n}} = \frac{1}{9}.$$
Hence, interval of convergence contains $(0,18)$. For $x=18$ we get $\sum \frac{(-1)^n}{n}$, which converges. For $x=0$ we get divergence. So the interval of convergence is $(0,18]$.