Chapter 8: Infinite Sequences and Series
Section 8.2: Series
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Example 8.2.16
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a)
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Show that Leibniz' theorem on the convergence of alternating series applies to the alternating harmonic series. (See Table 8.2.2.)
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b)
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Use Maple to show that the sequence of partial sums converges to .
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c)
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Test the claim that a partial sum is closer to the sum than the magnitude of the first neglected term.
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Solution
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Part (a)
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The alternating harmonic series is , with defining a sequence of positive terms that decrease monotonically to zero. Hence the conditions of Leibniz' theorem hold, and the sequence converges conditionally.
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Part (b)
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Obtain the sum of the series
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Write the alternating harmonic series.
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Context Panel: Evaluate and Display Inline
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Obtain an expression for the th partial sum
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Control-drag the series and change ∞ to .
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Context Panel: Evaluate and Display Inline
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Context Panel: Assign to a Name≻S[k]
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Display the first few partial sums
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Type and press the Enter key.
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Context Panel: Sequence≻
In the resulting dialog box, set to
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Obtain the limit of the partial sums
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Calculus palette: Limit template≻Apply to
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Context Panel: Evaluate and Display Inline
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Figure 8.3.16(a) shows the convergence of the first 15 members of the sequence of partial sums to .
>
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use plots in
module()
local Sk,X,Y,p1,p2,p3,k;
unassign('S');
Sk:=k->sum((-1)^(n+1)/n,n=1..k);
X:=[seq(k,k=1..15)];
Y:=[seq(Sk(k),k=1..15)];
p1:=pointplot(X,Y,symbol=solidcircle,symbolsize=15,color=red,labels=[k,typeset(S[k])],view=[0..15,0..1]);
p2:=plot(ln(2),k=0..15,color=black);
p3:=display(p1,p2);
print(p3)
end module:
end use:
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Figure 8.2.16(a) Convergence of to
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Part (c)
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Table 8.2.16(a) provides the evidence that a partial sum is closer to the sum than the magnitude of the first neglected term. The first column of the table lists , the order of the partial sum. The middle column lists , the measure of how far the partial sum is from the sum . The third column of the table is the value of , the magnitude of the first neglected term.
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module()
local SK,r,M,j;
SK:=sum((-1)^(n+1)/n,n=1..k);
for j from 1 to 10 do
r[j]:=[j,abs(evalf(ln(2)-value(eval(SK,k=j)))),evalf(1/(j+1))]
od:
M:=Matrix([seq(r[j],j=1..10)]);
print(M);
end module:
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Table 8.2.16(a) Values of compared to
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In every instance, .
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