Power Series — Question 1

Section 11-14 · Calculus pages 2201–2202

Show solutionHide solution

*Question 1 Find the center, radius of convergence, and interval of convergence of ∑n=0∞xn3n.\sum_{n=0}^{\infty}\frac{x^n}{3^n}. Test x=−3x=-3 and x=3x=3 separately and identify each resulting endpoint series.

*Question 1 – Solution

∑n=0∞xn3n=∑n=0∞(x3)n.\sum_{n=0}^{\infty}\frac{x^n}{3^n}=\sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n. It converges exactly when |x/3|<1|x/3|<1, or |x|<3|x|<3. Thus the center is 00 and R=3R=3.

  • At x=3x=3, the series is ∑1\sum1, which diverges.

  • At x=−3x=-3, the series is ∑(−1)n\sum(-1)^n. Its terms do not approach 00, so it diverges.

The interval is (−3,3)\boxed{(-3,3)}. Neither endpoint is included.