1. Using ratio test to find radius of convergence

f(x) = sigma n=0 to infinity of (2n/8^n)x^(3n)

1. Use the ratio test to find the radius of convergence of the power series.
2. Find f '(x) and write out the first three (non-zero) terms of the power series for f '(x).

1. I got

a(n) = (2n * x^(3n)) / 8^n
a(n + 1)= [2(n+1) * x^(3(n + 1))] / 8^(n + 1)

After that I did a(n+1) / a(n).

Simplified and in the end got (x^3 / 8) which resulted in (x^3 < 8) making it equal R = 2. I'm wandering if this is the correct answer or not?

2. As for this question. I got the derivative as.

= (2n/8^n) * 3nx^(3n-1)

f '(x) = (6n^2/8^n) * x^(3n-1)

Is that the correct derivative? Also, to find the first three terms, I just plug in numbers starting 0 and then add up the first 3 non-zero answers, correct?

Thanks.

2. Originally Posted by Kyeong
1. I got

a(n) = (2n * x^(3n)) / 8^n
a(n + 1)= [2(n+1) * x^(3(n + 1))] / 8^(n + 1)

After that I did a(n+1) / a(n).

Simplified and in the end got (x^3 / 8) which resulted in (x^3 < 8) making it equal R = 2. I'm wandering if this is the correct answer or not?
Correct!

$\frac{a_{n+1}}{a_n} = \frac{\frac{2(n + 1)x^{3n+3}}{8^{n+1}}}{\frac{2nx^{3n}}{8^n}} = \frac{2(n+1)x^{3n + 3}}{8^{n+1}}\cdot\frac{8^n}{2nx^{3n}}$

$= \frac{2(n+1)x^{3n + 3}}{2nx^{3n}}\cdot\frac{8^n}{8^{n+1}} = \left(1 + \frac1n\right)\frac{x^3}8$

which gives

$\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\ri ght\rvert = \frac{\left\lvert x^3\right\rvert}8.$

So to find the radius of convergence, we do

$\frac{\left\lvert x^3\right\rvert}8 < 1$

$\Rightarrow\left\lvert x^3\right\rvert < 8$

$\Rightarrow\left\lvert x\right\rvert^3 < 8$

$\Rightarrow\left\lvert x\right\rvert < 2$

$\Rightarrow-2 < x < 2$

Originally Posted by Kyeong
2. As for this question. I got the derivative as.

= (2n/8^n) * 3nx^(3n-1)

f '(x) = (6n^2/8^n) * x^(3n-1)

Is that the correct derivative? Also, to find the first three terms, I just plug in numbers starting 0 and then add up the first 3 non-zero answers, correct?

Thanks.
Yes, but don't forget that this is still a series:

$f'(x) = \sum_{n=1}^\infty\frac{6n^2x^{3n - 1}}{8^n}$

3. Oh yea, I almost forgot. Thanks for reminding me. I was doubting my answer and you just clarified what I thought. Thanks again.