Chapter 8: Infinite Sequences and Series
Section 8.2: Series
Example 8.2.11
Use Maple to sum the series ∑n=1∞1n n+2 and show that the sum is the limit of the sequence of partial sums.
Solution
Mathematical Solution
A partial fraction decomposition gives 1n n+1=121n−1n+2, so that the given series is a telescoping series. (See Table 8.2.2.)
Twice the kth partial sum is then
2 Sk
=11−13+12−14+13−15+⋯+1k−2−1k+1k−1−1k+1+1k−1k+2
=1+12−1k+1−1k+2
=32−1k+1+1k+2
where the red terms cancel in pairs, but the two bold terms will be left in any finite sum.
Consequently, the sum of the series is given by
limk→∞Sk=34−limk→∞121k+1+1k+2=34−0=34
Maple Solution
Obtain the sum of the series
Control-drag the series.
Context Panel: Evaluate and Display Inline
∑n=1∞1n n+2 = 34
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=1k1n n+2 = −12k+1−12k+2+34→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
−12k+1−12k+2+34
→sequence w.r.t. k
13,1124,2140,1730,2542,69112,91144,2945,3655,175264,209312,123182,143210,329480,375544
Obtain the limit of the partial sums
Calculus palette: Limit template≻Apply to Sk
limk→∞Sk = 34
Figure 8.2.11(a) shows the convergence of the first 15 members of the sequence of partial sums to S=3/4.
use plots in module() local Sk,X,Y,p1,p2,p3,k; unassign('S'); Sk:=k->sum(1/n/(n+2),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(3/4,k=0..15,color=black); p3:=display(p1,p2); print(p3) end module: end use:
Figure 8.2.11(a) Convergence of Sk to S=3/4
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