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Example 8.3.8 (b): Power Series?

Is the series f(x) = 32nxn a power series? If so, list center, radius of convergence, and general term an.

Yes, this is also a power series centered at c = 0. The general term here is

an = 32n = (32)n = 9n

According to our formula the radius of convergence is:

r = lim sup | an / an+1 | = lim sup | 9n / 9n+1 | = 1/9

Thus the series converges for |x| < 1/9. To determine exactly where the series converges we should check the endpoints of this interval manually:

  • For x=-1/9: 32nxn = 9n(-1/9)n = (-1)n, which diverges.
  • For x=1/9: 32nxn = 9n(1/9)n = 1n, which diverges as well.
32nxn f(x) = 1/1-9x
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