Power Series Math Example 4

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Example 4

medium
Differentiate ∑n=0∞xn=11−x\sum_{n=0}^{\infty} x^n = \frac{1}{1-x} term by term to find a new series identity.

Solution

  1. 1
    ddx∑n=0∞xn=∑n=1∞nxn−1\frac{d}{dx}\sum_{n=0}^{\infty}x^n = \sum_{n=1}^{\infty}nx^{n-1}.
  2. 2
    ddx11−x=1(1−x)2\frac{d}{dx}\frac{1}{1-x} = \frac{1}{(1-x)^2}.
  3. 3
    Therefore ∑n=1∞nxn−1=1(1−x)2\sum_{n=1}^{\infty}nx^{n-1} = \frac{1}{(1-x)^2} for ∣x∣<1|x|<1.

Answer

∑n=1∞nxn−1=1(1−x)2,∣x∣<1\sum_{n=1}^{\infty}nx^{n-1} = \frac{1}{(1-x)^2}, \quad |x|<1
Term-by-term differentiation is valid within the radius of convergence and produces a new closed-form identity.

About Power Series

An infinite series of the form ∑n=0∞an(x−c)n=a0+a1(x−c)+a2(x−c)2+⋯\sum_{n=0}^{\infty} a_n(x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots where cc is the center and ana_n are the coefficients. A power series defines a function of xx wherever it converges.

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More Power Series Examples

Example 1 medium

Find the radius of convergence of [formula].

Example 2 hard

Find the interval of convergence of [formula].

Example 3 easy

Find the radius of convergence of [formula].

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