Power Series — Question 4

PDF ↗

Question 4

Find the radius and interval of convergence of ∑n=1∞xn/n2\displaystyle\sum_{n=1}^{\infty}x^n/n^2. Explain why its radius matches Question 3 while both endpoints now converge.

Original worksheet page 1: question and worked solution for 4-14-004
Show solutionHide solution

Question 4 – Solution

Step 1: Find the radius.

|xn+1/(n+1)2xn/n2|=|x|(nn+1)2→|x|.\left|\frac{x^{n+1}/(n+1)^2}{x^n/n^2}\right|=|x|\left(\frac{n}{n+1}\right)^2\longrightarrow|x|. Thus the interior condition is |x|<1|x|<1 and R=1R=1.

Step 2: Test the endpoints.
  • At x=1x=1, the series is ∑1/n2\sum1/n^2, which converges.

  • At x=−1x=-1, the absolute-value series is again ∑1/n2\sum1/n^2, so the series converges absolutely.

Conclusion.

The interval is [−1,1]\boxed{[-1,1]}. Both this series and ∑xn/n\sum x^n/n have coefficient roots tending to 11, hence the same radius, but the stronger n−2n^{-2} decay makes both boundary series absolutely convergent.

Original worksheet page 2: question and worked solution for 4-14-004

Original worksheet layout. Use Enlarge or open the PDF for a closer view.