Power Series — Question 2

PDF ↗

Question 2

Find the radius and interval of convergence of ∑n=1∞nxn4n\displaystyle\sum_{n=1}^{\infty}\frac{n x^n}{4^n}. Explain why the polynomial factor nn does not change the radius but does determine the endpoint behavior.

Original worksheet page 1: question and worked solution for 4-14-002
Show solutionHide solution

Question 2 – Solution

Step 1: Use the Ratio Test.

For an=nxn/4na_n=nx^n/4^n, |an+1an|=n+1n|x|4→|x|4.\left|\frac{a_{n+1}}{a_n}\right|=\frac{n+1}{n}\frac{|x|}{4}\longrightarrow\frac{|x|}{4}. Convergence requires |x|/4<1|x|/4<1, so |x|<4|x|<4 and R=4R=4.

Step 2: Test the endpoints.
  • At x=4x=4, the terms are nn, which do not approach 00.

  • At x=−4x=-4, the terms are n(−1)nn(-1)^n, whose magnitudes grow.

Both endpoint series diverge by the nth-term test.

Conclusion.

The interval is (−4,4)\boxed{(-4,4)}. The factor (n+1)/n→1(n+1)/n\to1, so nn does not alter the radius; at the endpoints it prevents the terms from tending to zero.

Original worksheet page 2: question and worked solution for 4-14-002

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