Chapter 8: Infinite Sequences and Series
Section 8.2: Series
Example 8.2.1
Sum the series∑n=0∞1/3n and show that the sum is the limit of the sequence of partial sums.
Solution
Mathematical Solution
The geometric series ∑n=0∞rn sums to S=11−r, and the finite sum Sk=∑n=0krn adds to 1−rk+11−r .
Since the given series is a geometric series with r=1/3, its sum is
11−1/3=33−1=32
The partial sum up through n=k is Sk=1−1/3k+11−1/3 , which, in the limit as k→∞, becomes 1−0 1−1/3=32.
Figure 8.2.1(a) shows the first few partial sums rapidly converging to S=3/2.
use plots in module() local Sk,X,Y,p1,p2,p3,k; Sk:=k->(1-(1/3)^(k+1))/(1-1/3); X:=[seq(k,k=0..10)]; Y:=[seq(Sk(k),k=0..10)]; p1:=pointplot(X,Y,symbol=solidcircle,symbolsize=15,color=red,labels=[k,typeset(S[k])],view=[0..10,0..2]); p2:=plot(1.5,k=0..10,color=black); p3:=display(p1,p2); print(p3) end module: end use:
Figure 8.2.1(a) Convergence of Sk to S
Maple Solution
Maple "knows" how to sum a geometric series:
Obtain the sum of the series
Control-drag the given series.
Context Panel: Evaluate and Display Inline
∑n=0∞1/3n = 32
Obtain Sk and the first few partial sums
∑n=0k13n = −313k+12+32→sequence w.r.t. k1,43,139,4027,12181,364243,1093729,32802187,98416561,2952419683,8857359049
Obtain the limit of the sequence of partial sums
Calculus palette: Limit operator
Expression palette: Summation template Sum the series to n=k.
limk→∞∑n=0k1/3n = 32
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