Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
Example 8.4.2
Determine the radius of convergence and the interval of convergence for the power series ∑n=0∞xnn+3.
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→∞an/an+1 = limn→∞n+4n+3=1
At the right endpoint x=1, the given power series becomes Σ an , which diverges by part (1) of the Limit-Comparison test if the comparison series is taken as the divergent harmonic series Σ 1/n.
At the left endpoint x=−1, the given power series becomes the alternating series Σ −1nan , which converges conditionally by the Leibniz test: 1/n+3 decreases monotonically to zero.
Hence, the interval of convergence is −R,R=−1,1.
Maple Solution
Define the general coefficient an as a function of n
Write an=… Context Panel: Assign Function
an=1n+3→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 = 1
Limit-Comparison test at x=R=1
Calculus palette: Limit template Context Panel: Evaluate and Display Inline
limn→∞an1/n = 1
The comparison series is the divergent harmonic series Σ 1/n, so by part (1) of the Limit-Comparison test, the given series diverges at x=R=1.
At x=−R=−1, the series is alternating, and converges conditionally by the Leibniz test because an−1n = 1/n+3, which tends monotonically to zero as n→∞.
Maple can actually sum this series, and gives
∑n=0∞xnn+3 = −ln1−xx3−1x2−12x
for the sum, provided appropriate assumptions are imposed on x .
Figure 8.4.2(a) is a graph of this function on the interval of convergence.
Figure 8.4.2(a) Graph of the sum of the series
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