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?