Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
Example 8.4.5
Determine the radius of convergence and the interval of convergence for the power series ∑n=1∞xnn2.
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→∞1/n21/n+12=limn→∞n+1n2=1
At the right endpoint x=R=1, the given power series becomes Σ an Rn=Σ 1/n2, which converges absolutely because it is the convergent p-series where p=2>1.
At the left endpoint x=−R=−1, the given power series becomes the alternating series Σ −1n/n2, which has just been shown to converge absolutely. Of course, since 1/n2 decreases monotonically to zero, the Leibniz test would establish at least conditional convergence.
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=1n2→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
At the right endpoint where x=R=1, the series becomes the absolutely convergent p-series, with p=2>1.
At the left endpoint where x=R=−1, the series becomes the alternating series Σ −1n/n2, which converges absolutely because the series of absolute values is just the same series examined at the right endpoint.
Maple can actually sum this series in terms of the special function polylog, and gives
∑n=1∞xnn2 = polylog2,x
for the sum, if it is assumed that x≤1.
Figure 8.4.5(a) is a graph of this function on the interval of convergence.
Figure 8.4.5(a) Graph of the sum of the series
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