Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
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Example 8.4.12
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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 diverges by part (3) of the Limit-Comparison test if the comparison series is taken as the divergent series , the harmonic series in disguise.
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The relevant calculation that must be made is . Since the comparison series diverges, so too does the power series at the right endpoint .
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At the left endpoint , the given power series becomes the alternating series , which converges conditionally by the Leibniz test, once it is noted that decreases monotonically to zero as .
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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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Test for convergence at
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Write
Context Panel: Assign Function
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Calculus palette: Limit template
Context Panel: Evaluate and Display Inline
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=
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The comparison series is a disguised form of the divergent harmonic series. By part (3) of the Limit-Comparison test, since the comparison series diverges, so too does the power series at the right endpoint .
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At the left endpoint , the given power series becomes the alternating series , which converges conditionally by the Leibniz test, once it is noted that decreases monotonically to zero as .
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Hence, the interval of convergence is .
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