Advanced High-School Mathematics

286 CHAPTER 5 Series and Differential Equations

In the examples above we computed intervals within which we are
guaranteedconvergence of the power series. Next, note that for values
ofx outside the radius of convergence wecannot have convergence,
for then the limit of the ratios will be greater than 1, preventing the
individual terms from approaching 0. This raises the question ofcon-
vergence at the endpoints. We shall investigate this in the examples
already considered above.

Example 6. We have seen that the power series

∑∞

n=0

(−1)nxn

2 n+ 1

has

radius of convergenceR= 1 meaning that this series will converge in
the interval − 1 < x < 1. What ifx = −1? What if x = 1? Well,
we can take these up separately, using the methods of the previous two
subsections. Ifx=−1, we have the series

∑∞

n=0

(−1)n(−1)n

2 n+ 1

=

∑∞

n=0

1

2 n+ 1

,

whichdiverges (use the Limit Comparison Test against the har-
monic series). Ifx= 1, then the series becomes

∑∞

n=0

(−1)n

2 n+ 1

,

which converges by theAlternating Series Test. We therefore know
the full story and can state theinterval of convergence:

∑∞

n=0

(−1)nxn

2 n+ 1

has interval of convergence − 1 < x≤ 1.

Example 7. We saw that the power series

∑∞

n=0

nxn

2 n

has radius of con-

vergencer = 2. What about the behavior of the series whenx=±2.
Ifx=−2 we have the series

∑∞

n=0

n(−2)n

2 n

=

∑∞

n=0

(−1)nn which diverges,

whereas whenx= 2, we have