Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
Example 8.4.24
Determine the radius of convergence and the interval of convergence for the power series ∑n=1∞1⋅3⋅⋯⋅2 n−12⋅5⋅⋯⋅3 n−1xn.
Even though (7) in Table 8.4.1 claims that absolute convergence at one end of the interval of convergence implies absolute convergence at the other, if the convergence at an endpoint is absolute, verify that it also absolute at the other.
Solution
Mathematical Solution
Since the given power series contains the powers xn, the radius of convergence is given by
R=limn→∞anan+1 = limn→∞1⋅3⋅⋯⋅2 n−12⋅5⋅⋯⋅3 n−12⋅5⋅⋯⋅3 n+1−11⋅3⋅⋯⋅2 n+1−1 = limn→∞3 n+22 n+1=32
At the right endpoint x=3/2, the given power series becomes Σ an 3/2n, which diverges by part (3) of the Limit-Comparison test if the comparison series is taken as the divergent harmonic series Σ 1/n. In the limit as n→∞, an 3/2n1/n→∞, as will be shown in the Maple Solution, below.
At the left endpoint −3/2, the given power series becomes the alternating series Σ −1nan 3/2n, which converges conditionally by the Leibniz test: an3/2n decreases monotonically to zero, as will be shown in the Maple Solution, below.
Hence, the interval of convergence is −R,R=−3/2,3/2.
Maple Solution
The key to making the calculations in this solution tractable is the representation of the products of integers with the product operator, available in the Expression palette. Maple evaluates these products in terms of the gamma function, the generalization of the factorial function. Without these tools, these calculations would be exceedingly difficult.
Define the general coefficient an as a function of n
Expression palette: Product template Write an=… Context Panel: Assign Function
an=∏k=1n2 k−1∏k=1n3 k−1→assign as functiona
Obtain the radius of convergence
Calculus palette: Limit template Context Panel: Assign Name
R=limn→∞anan+1→assign
Display R, the radius of convergence
Write R Context Panel: Evaluate and Display Inline
R = 32
Limit-Comparison test at x=R=3/2
Calculus palette: Limit template Context Panel: Evaluate and Display Inline
limn→∞an⋅3/2n1/n = ∞
Limiting value of the general term of the series at x=−R=−3/2
limn→∞an⋅3/2n = 0
At the right endpoint x=3/2, the given power series becomes Σ an 3/2n, which diverges by part (3) of the Limit-Comparison test if the comparison series is taken as the divergent harmonic series Σ 1/n. In the limit as n→∞, an 3/2n1/n→∞.
At the left endpoint −3/2, the given power series becomes the alternating series Σ −1nan 3/2n, which converges conditionally by the Leibniz test: an3/2n decreases monotonically to zero, as shown in Figure 8.4.24(a).
module() local a,p; a:=(product(2*k-1,k=1..n))/(product(3*k-1, k=1..n)); p:=plot(a*(3/2)^n,n=1..20,labels=[x,f],tickmarks=[10,3]); print(p); end module:
Figure 8.4.24(a) Graph of fx=ax 3/2x
plot(x*hypergeom([1,3/2], [5/3],2*x*(1/3)),x=-3/2.. 3/2,-1..20,labels=[x,y], tickmarks=[3,3],size=[200,200]);
Figure 8.4.24(b) Graph of the sum of the series
Maple sums this series in terms of the special function hypergeom, and obtains
∑n=1∞an⋅xn assuming x<3/2 = 12xhypergeom1,32,53,23x
for the sum.
Figure 8.4.24(b) is a graph of this function on the interval of convergence.
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