Power Series and Functions — Question 2

Section 11-15 · Calculus pages 2223–2224

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*Question 2 Obtain a power series centered at 00 for f(x)=1/(1+x)f(x)=1/(1+x) by substituting into the geometric series. State its radius and interval of convergence and test both endpoints directly.

*Question 2 – Solution

Replace xx by −x-x in 1/(1−x)=∑xn1/(1-x)=\sum x^n: 11+x=∑n=0∞(−x)n=∑n=0∞(−1)nxn.\boxed{\frac1{1+x}=\sum_{n=0}^{\infty}(-x)^n=\sum_{n=0}^{\infty}(-1)^n x^n}.

The geometric condition is |−x|<1|-x|<1, equivalent to |x|<1|x|<1, so R=1R=1.

  • At x=1x=1, the series is ∑(−1)n\sum(-1)^n, which diverges by the nth-term test.

  • At x=−1x=-1, it is ∑1\sum1, which diverges.

The interval is (−1,1)\boxed{(-1,1)}. Although 1/(1+x)1/(1+x) is finite at x=1x=1, its Taylor series about 00 does not converge there.