Chapter 8: Infinite Sequences and Series
Section 8.3: Convergence Tests
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Example 8.3.18
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Determine if the series diverges, converges absolutely, or converges conditionally.
If it converges conditionally, determine if it also converge absolutely.
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Solution
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Mathematical Solution
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The series is alternating, suggesting the Leibniz test. For this, must eventually become a monotone decreasing sequence with limit zero. Consequently, the following calculations are executed.
= and for
The conditions of the Leibniz test apply, and by that test, the series converges conditionally.
In Example 8.3.16, the Integral test is applied to the sequence , giving
=
Since this integral diverges, the given series does not converge absolutely.
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Maple Solution
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Figure 8.3.18(a) contains a graph of the function (in red) and of its derivative (in green).
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On the basis of this graph, it may be conjectured that is monotone decreasing and bounded below by zero, provided . (The derivative appears to be negative for .)
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Consequently, the Integral test may be tried as a test for absolute convergence, provided the integration starts from, say, .
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>
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module()
local F,p,N;
F:=ln(x)/x;
N:=10;
p:=plot([F,diff(F,x)],x=1..N,color=[red,green],view=[0..N,-1/10..1/2],tickmarks=[N,4],labels=[x,y]);
print(p);
end module:
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Figure 8.3.18(a) Graph of (red) and (green)
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Calculus palette: Definite integral template
Context Panel: Evaluate and Display Inline
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=
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Since the integral diverges, the series does not converge absolutely.
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The following two calculations support the use of the Integral test.
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Calculus palette: Limit operator
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Context Panel: Evaluate and Display Inline
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=
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Calculus palette: Differentiation operator
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Context Panel: Evaluate and Display Inline
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=
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The derivative is negative for . (This is consistent with Figure 8.3.18(a).) Since the criteria of the test are satisfied, it follows that the series does not converge absolutely. The Leibniz test applies, and by it, this alternating series is seen to converge conditionally.
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