Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
Example 8.4.3
Determine the radius of convergence and the interval of convergence for the power series ∑n=1∞−1n+1xnn5.
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+15n5=1
At the right endpoint x=R=1, the given power series becomes the alternating series Σ −1n+1/n5, which converges conditionally by the Leibniz test since 1/n5 decreases monotonically to zero.
At the left endpoint x=−R=−1, the given power series becomes the divergent p-series −Σ 1/n1/5 because p=1/5<1.
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+1n5→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
Test for convergence at x=R=−1
Write the general term of the series at x=−1. Press the Enter key.
Context Panel: Simplify≻Assuming Integer The result is the general term of a divergent p-series because p=1/5<1.
an⋅−Rn
−1n+1−1nn1/5
→assuming integer
−1n1/5
At the right endpoint x=R=1, the given power series becomes an alternating series in with 1/n5 tends monotonically to zero. By the Leibniz test, this series converges conditionally.
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