Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
Determine the radius of convergence and the interval of convergence for the power series ∑n=1∞n 3n xn.
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.
Since the given power series contains the powers xn, the radius of convergence is given by
R=limn→∞an/an+1 = limn→∞n⋅3nn+1 3n+1=limn→∞n3n+1=1/3
At the right endpoint x=R=1/3, the given power series becomes Σ an Rn=Σ n, which diverges by the nth-term test.
At the left endpoint x=−R=−1/3, the given power series becomes the alternating series Σ −1nn, which also diverges by the nth-term test.
Hence, the interval of convergence is −R,R=−1/3,1/3.
Define the general coefficient an as a function of n
Context Panel: Assign Function
an=n⋅3n→assign as functiona
Obtain the radius of convergence
Calculus palette: Limit template
Context Panel: Assign Name
Display R, the radius of convergence
Context Panel: Evaluate and Display Inline
R = 13
Test for convergence at x=R=1/3
Write an⋅Rn and press the Enter key.
Context Panel: Simplify≻Assuming Integer
Test for convergence at x=R=−1/3
Write an⋅−Rn and press the Enter key.
At the right endpoint x=1/3, the power series becomes Σ n, which, by the nth-term test, diverges.
At the left endpoint x=−1/3, the power series becomes Σ −1nn, which, by the nth-term test, diverges.
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