Interval of Convergence and radius of convergence

[Ohoneo]

I'm supposed to find both of these for the summation of x^(2k+1)/3^(k-1).

I used the ratio test, and got x^(2k+2)3^k * 3^(k-1)/x^(2k+1) which simplified to be x3^(k-1)/3^k.
The 3^k-1/3^k simplifies to be 1/3, so you get x/3, which you set <1, which gives you an interval of convergence of -3<x<3.
Now, I'm stuck from there. The answer to the problem is R = sqrt(3) and the interval is (-sqrt(3), sqrt(3)).

 

[soroban]

Hello, Ohoneo!

Your algebra is off . . .

I'm supposed to find both of these for the summation of: .\(\displaystyle \dfrac{x^{2k+1}}{3^{k-1}}\)

The ratio is: .\(\displaystyle R \:=\:\dfrac{x^{2k+3}}{3^k}\cdot\dfrac{3^{k-1}}{x^{2k+1}} \;=\;\dfrac{x^2}{3}\)

We want: .\(\displaystyle \displaystyle \left|\frac{x^2}{3}\right| \:<\:1 \quad\Rightarrow\quad |x^2| \:<\:3 \quad\Rightarrow\quad |x| \:<\:\sqrt{3}\)


The interval of convergence is: .\(\displaystyle (\text{-}\sqrt{3},\:\sqrt{3})\)

The radius of convergence is: .\(\displaystyle r \:=\:\sqrt{3}\)

 

[Ohoneo]

This all made it a lot clearer, thank you! I just have one question, in the first step (when applying the ratio step) why does it becomes x^2k+3 on top? I know you add one to k, so I was under the impression it would be x^2k+2. Thanks for your help!