Chapter 8: Infinite Sequences and Series
Section 8.2: Series
Example 8.2.4
Sum the series ∑n=1∞1n n+1 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=1n−1n+1, so that the given series is a telescoping series. (See Table 8.2.2.)
The kth partial sum is then
Sk
=11−12+12−13+13−14+⋯+1k−1−1k+1k−1k+1
=1−1k+1
Consequently, the sum of the series is given by
limk→∞Sk=1−limk→∞1k+1=1−0=1
Maple Solution
Obtain the sum of the series
Control-drag the series.
Context Panel: Evaluate and Display Inline
∑n=1∞1n n+1 = 1
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+1 = −1k+1+1→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
−1k+1+1
→sequence w.r.t. k
12,23,34,45,56,67,78,89,910,1011,1112,1213,1314,1415,1516
Obtain the limit of the partial sums
Calculus palette: Limit template≻Apply to Sk
limk→∞Sk = 1
Figure 8.2.4(a) shows the convergence of the first 15 members of the sequence of partial sums to S=1.
use plots in module() local Sk,X,Y,p1,p2,p3,k; unassign('S'); Sk:=k->sum(1/n/(n+1),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(1,k=0..15,color=black); p3:=display(p1,p2); print(p3) end module: end use:
Figure 8.2.4(a) Convergence of Sk to S=1
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