Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
Example 8.4.23
Determine the radius of convergence and the interval of convergence for the power series ∑n=1∞n22n4 x+5n.
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 4 x+5n, the radius of convergence is given by
R=14limn→∞an/an+1 = 14limn→∞n2/2nn+12/2n+1=14limn→∞2nn+12=24=12
The endpoints of the interval of convergence are then x=−5/4 ±1/2, or −7/4 and −3/4.
At the right endpoint x=−3/4, 4 x+5=2, so the given power series becomes Σ n2, which diverges by the nth-term test.
At the left endpoint x=−7/4, 4 x+5=−2, so the given power series becomes the alternating series Σ −1n n2, which also diverges by the nth-term test.
Hence, the interval of convergence is −7/4,−3/4.
Maple Solution
Define the general coefficient an as a function of n
Write an=… Context Panel: Assign Function
an=n22n→assign as functiona
Obtain the radius of convergence
Calculus palette: Limit template Context Panel: Assign Name
R=14 limn→∞anan+1→assign
Display R, the radius of convergence
Write R Context Panel: Evaluate and Display Inline
R = 12
Assuming x∈−7/4,−3/4, Maple sums this series to
∑n=1∞n22n4 x+5n = −24x+74x+54x+33
Figure 8.4.23(a) is a graph of this function on the interval of convergence.
Figure 8.4.23(a) Graph of the sum of the series
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