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Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
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Example 8.4.21
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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 the nth-term test because .
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At the left endpoint , the given power series becomes the alternating series , which also diverges by the nth-term test because .
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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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Calculus palette: Limit template
Context Panel: Evaluate and Display Inline
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At the right endpoint , the given power series becomes , which diverges by the nth-term test because .
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At the left endpoint , the given power series becomes the alternating series , which also diverges by the nth-term test because .
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Hence, the interval of convergence is .
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Maple can actually sum this series - Figure 8.4.21(a) is a graph of this function on the interval of convergence.
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Enter the series.
Context Panel: Simplify≻Assuming Real Range≻
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>
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module()
local S,p;
S:=sum((n!)^2/(2*n)!*x^n,n=2..infinity) assuming abs(x)<4;
p:=plot(S,x=-4..4,0..20,labels=[x,y]);
print(p);
end module:
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Figure 8.4.21(a) Graph of the sum of the series
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