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Theorem: Maclaurin Series of 1/(1 - x)

The Maclaurin series of f(x)=11−xf(x) = \dfrac{1}{1-x} is ∑n=0∞xn\displaystyle\sum_{n=0}^{\infty} x^n, and

11−x=∑n=0∞xn=1+x+x2+x3+⋯for ∣x∣<1,\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots \qquad \text{for } |x| < 1,

while the series diverges for ∣x∣≥1|x| \ge 1.

26 viewsMathematics · Single Variable Calculus
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