Chapter 8: Infinite Sequences and Series
Section 8.2: Series
Example 8.2.16
Show that Leibniz' theorem on the convergence of alternating series applies to the alternating harmonic series. (See Table 8.2.2.)
Use Maple to show that the sequence of partial sums converges to ln2.
Test the claim that a partial sum is closer to the sum than the magnitude of the first neglected term.
Solution
Part (a)
The alternating harmonic series is ∑n=1∞−1n+1n, with an=1/n defining a sequence of positive terms that decrease monotonically to zero. Hence the conditions of Leibniz' theorem hold, and the sequence converges conditionally.
Part (b)
Obtain the sum of the series
Write the alternating harmonic series.
Context Panel: Evaluate and Display Inline
∑n=1∞−1n+1n = ln2
Obtain an expression for the kth partial sum
Control-drag the series and change ∞ to k.
Context Panel: Assign to a Name≻S[k]
∑n=1k−1n+1n = ln2−12−1kΨ12k+1−Ψ12k+12→assign to a nameSk
Display the first few partial sums
Type Sk and press the Enter key.
Context Panel: Sequence≻k In the resulting dialog box, set k=1 to k=15
Sk
ln2−12−1kΨ12k+1−Ψ12k+12
→sequence w.r.t. k
1,12,56,712,4760,3760,319420,533840,18792520,16272520,2041727720,1810727720,263111360360,237371360360,5227972072
Obtain the limit of the partial sums
Calculus palette: Limit template≻Apply to Sk
limk→∞Sk = ln2
Figure 8.3.16(a) shows the convergence of the first 15 members of the sequence of partial sums to S=ln2.
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:
Figure 8.2.16(a) Convergence of Sk to S=ln2
Part (c)
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 k, the order of the partial sum. The middle column lists S−Sk , the measure of how far the partial sum Sk is from the sum S. The third column of the table is the value of 1/k+1, the magnitude of the first neglected term.
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:
Table 8.2.16(a) Values of S−Sk compared to ak+1=1/n+1
In every instance, |S−Sk|≤ak+1=1/k+1.
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