Discussion
Taylor series of \(f(x) = \dfrac{1}{1-x}\) about \(x = 0\) is given as \(TS|_{f} = 1 + x + x^{2} + x^{3} + \ldots\). This series can be used to evaluate \(f(x)\) for
Answer: C
The series can be used for \(|x| \lt 1\).
\(1 + x + x^{2} + x^{3} + \ldots\) is a geometric series with common ratio x. It converges, and its sum equals \(1/(1 - x)\), exactly when \(|x| \lt 1\); its radius of convergence is 1, the distance from 0 to the singularity at \(x = 1\).
\(|x| \neq 1\) wrongly includes values such as x = 2, where the terms grow without bound. For \(x \lt -1\) the series also diverges. \(-1 \le x \lt 1\) wrongly includes \(x = -1\), where the series becomes 1 - 1 + 1 - 1 ..., which oscillates and has no sum, even though f(-1) = 1/2.
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