Chapter 8: Infinite Sequences and Series
Section 8.2: Series
Example 8.2.5
Use Maple to sum the series ∑n=3∞4n2−4 and show that the sum is the limit of the sequence of partial sums.
Solution
Obtain the sum of the series
Control-drag the series.
Context Panel: Evaluate and Display Inline
∑n=3∞4n2−4 = 2512
Obtain a general expression for the kth partial sum
Control-drag the series and change the upper limit of the sum from ∞ to k.
Context Panel: Assign to a Name≻S[k]
∑n=3k4n2−4 = −1k−1−1k−1k+1−1k+2+2512→assign to a nameSk
Display the first few partial sums
Write Sk and press the Enter key.
Context Panel: Sequence≻k In the dialog box that appears, set k=3 to k=15
Sk
−1k−1−1k−1k+1−1k+2+2512
→sequence w.r.t. k
45,1715,139105,1217840,775504,40432520,65593960,1681990,24771430,35252002,48732730,3943921840,5209128560
Obtain the limit of the partial sums
Calculus palette: Limit Operator≻Apply to Sk
limk→∞Sk = 2512
Figure 8.2.5(a) shows the convergence of the first 15 members of the sequence of partial sums to S=25/12.
use plots in module() local SK,X,Y,p1,p2,p3,k; unassign('S'); SK:=k->sum(4/(n^2-4),n=3..k); X:=[seq(k,k=3..15)]; Y:=[seq(SK(k),k=3..15)]; p1:=pointplot(X,Y,symbol=solidcircle,symbolsize=15,color=red,labels=[k,typeset(S[k])],view=[0..15,0..2.5]); p2:=plot(25/12,k=0..15,color=black); p3:=display(p1,p2); print(p3) end module: end use:
Figure 8.2.5(a) Convergence of Sk to S=25/12
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