The series Σ(x/3)^n (from n=0 to ∞) is a geometric power series. For which values of x does it converge?
AAll real x
B|x| < 1
C|x| < 3
Dx = 0 only
This is a geometric series with ratio r = x/3. It converges when |r| < 1, i.e., |x/3| < 1, i.e., |x| < 3. The radius of convergence is 3, not 1. A common error is confusing the series with the standard Σx^n and concluding |x| < 1, ignoring that the ratio is x/3.
Question 2 True / False
A power series that converges for all x in (-5, 5) can be differentiated term by term, and the resulting power series also converges for all x in (-5, 5).
TTrue
FFalse
Answer: True
Term-by-term differentiation of a power series is valid inside its interval of convergence, and the resulting series has the same radius of convergence. Endpoint behavior may change, but the open interval (-5, 5) is preserved. This is one of the powerful properties that makes power series so useful — they behave like polynomials inside their interval of convergence.
Question 3 Short Answer
Starting from the geometric series 1/(1-x) = Σx^n for |x| < 1, how would you find a power series for 1/(1-x²)?
Think about your answer, then reveal below.
Model answer: Substitute x² for x: 1/(1-x²) = Σ(x²)^n = Σx^(2n) for |x²| < 1, i.e., |x| < 1.
Substituting x² into the geometric series is valid as long as the new ratio |x²| < 1, which gives |x| < 1. The resulting series Σx^(2n) = 1 + x² + x⁴ + x⁶ + ... converges on the same interval. This technique — deriving new series by substituting into known ones — is far faster than computing coefficients directly.