Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
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Example 8.4.5
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Determine the radius of convergence and the interval of convergence for the power series .
Even though (7) in Table 8.4.1 claims that absolute convergence at one end of the interval of convergence implies absolute convergence at the other, if the convergence at an endpoint is absolute, verify that it also absolute at the other.
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Solution
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Mathematical Solution
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Since the given power series contains the powers , the radius of convergence is given by
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At the right endpoint , the given power series becomes , which converges absolutely because it is the convergent p-series where .
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At the left endpoint , the given power series becomes the alternating series , which has just been shown to converge absolutely. Of course, since decreases monotonically to zero, the Leibniz test would establish at least conditional convergence.
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Hence, the interval of convergence is .
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Maple Solution
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Define the general coefficient as a function of
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Write
Context Panel: Assign Function
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Obtain the radius of convergence
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Calculus palette: Limit template
Context Panel: Assign Name
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Display , the radius of convergence
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Write
Context Panel: Evaluate and Display Inline
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At the right endpoint where , the series becomes the absolutely convergent p-series, with .
At the left endpoint where , the series becomes the alternating series , which converges absolutely because the series of absolute values is just the same series examined at the right endpoint.
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Hence, the interval of convergence is .
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Maple can actually sum this series in terms of the special function polylog, and gives
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for the sum, if it is assumed that .
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Figure 8.4.5(a) is a graph of this function on the interval of convergence.
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Figure 8.4.5(a) Graph of the sum of the series
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