Power Series Math Example 2

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Example 2

hard
Find the interval of convergence of ∑n=1∞(−1)nxnn\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n}.

Solution

  1. 1
    Ratio test: L=∣x∣<1L = |x| < 1, so R=1R=1.
  2. 2
    x=1x=1: ∑(−1)n/n\sum(-1)^n/n — alternating harmonic, converges.
  3. 3
    x=−1x=-1: ∑1/n\sum 1/n — harmonic, diverges.
  4. 4
    Interval: (−1,1](-1, 1].

Answer

Interval of convergence: (−1,1](-1, 1].
Always check endpoints individually. At x=1x=1 the series converges; at x=−1x=-1 it diverges.

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.

Learn more about Power Series →

More Power Series Examples

Example 1 medium

Find the radius of convergence of [formula].

Example 3 easy

Find the radius of convergence of [formula].

Example 4 medium

Differentiate [formula] term by term to find a new series identity.

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