What Is the Radius and Interval of Convergence for This Series?

[stunner5000pt]

Homework Statement

Find the radius of convergence and interval of convergence for the following infinite series
$$\sum_{n=1}^{\∞} \frac{x^n n^2}{3 \cdot 6 \cdot 9 \cdot ... (3n)}$$

Homework Equations

Ratio test

The Attempt at a Solution

Using ratio test we get
im not sure how to put absolute value signs but
$$\frac{(n+1)^2 x^{n+1}}{3 \cdot 6 \cdot 9 ... (3n) \cdot 3(n+1)} \frac{3 \cdot 6 \cdot 9 \cdot 3n}{n^2 x^n}$$

and this becomes
$$\frac{(n+1)^2 x}{3 n^2 (n+1)}$$
and that simplifies to

$$\frac{x(n+1)}{3n^2}$$

now here is where I have the trouble. the bottom of the fraction above is 'stronger' than the top which means that when we put the above < 1, it does not solve.

Can you please check if I did all the math correctly? Your assistance is greatly appreciated!

 

[jbunniii]

$$\frac{x(n+1)}{3n^2}$$

Yes, this looks right. Now ignore the $x$ for a moment, and consider this limit:
$$\lim_{n \rightarrow \infty} \frac{n+1}{3n^2}$$
What does this equal?

 

[LCKurtz]

You should have absolute values:$$
\lim_{n \rightarrow \infty}\left| \frac {x(n+1)}{3n^2}\right |$$
What do you get for that limit and what does it tell you?

 

[stunner5000pt]

You should have absolute values:$$
\lim_{n \rightarrow \infty}\left| \frac {x(n+1)}{3n^2}\right |$$
What do you get for that limit and what does it tell you?

thanks for your reply.

Isn't the limit zero? Wouldnt that mean that we can't make the limit <1?

 

[LCKurtz]

thanks for your reply.

Isn't the limit zero? Wouldnt that mean that we can't make the limit <1?

In my book, 0 < 1. What does that tell you about the series?

 

[stunner5000pt]

In my book, 0 < 1. What does that tell you about the series?

Ok that's perfect. that tells me that series converges.
How would I find the radius of convergence, though?

 

[LCKurtz]

Ok that's perfect. that tells me that series converges.
How would I find the radius of convergence, though?

Well, for what values of x does it converge?

 

[stunner5000pt]

Well, for what values of x does it converge?

it would converge for all x?

Would that mean the interval of convergence is -∞ to +∞ ?
Does that mean the radius is infinity?

 

[LCKurtz]

Since the limit ratio was zero no matter what value x has, the answer to all three questions is yes.

 

[stunner5000pt]

Since the limit ratio was zero no matter what value x has, the answer to all three questions is yes.

Thank you for your help