Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
Example 8.4.8
Determine the radius of convergence and the interval of convergence for the power series ∑n=1∞xnn 3n.
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→∞1n 3n1n+1 3n+1 = limn→∞3 n+1n=3
At the right endpoint x=R=3, the given power series becomes Σ an Rn=Σ 1/n, which is the divergent harmonic series.
At the left endpoint x=−R=−3, the given power series becomes the negative of the alternating harmonic series Σ −1n/n, which converges conditionally to −ln2.
Hence, the interval of convergence is −R,R=−3,3.
Maple Solution
Define the general coefficient an as a function of n
Write an=… Context Panel: Assign Function
an=1n 3n→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 = 3
Test for convergence at x=R=3
Write an⋅Rn Context Panel: Evaluate and Display Inline
an⋅Rn = 1n
Test for convergence at x=−R=−3
Write an⋅−Rn Context Panel: Evaluate and Display Inline
Context Panel: Simplify≻Assuming Integer
an⋅−Rn = −3nn3n= →assuming integer−1nn
At the right endpoint, the power series becomes the divergent harmonic series Σ 1/n.
At the left endpoint, the ower series becomes the negative of the conditionally convergent alternating harmonic series whose sum is −ln2.
Maple can actually sum this series, obtaining
∑n=1∞xnn 3n = = ln3−ln3−x
for the sum if x is assumed to lie in −3,3.
Figure 8.4.8(a) is a graph of this function on the interval of convergence.
Figure 8.4.8(a) Graph of the sum of the series
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