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8.6. POWER SERIES
of differentiating or integrating a series term by term to find new series.
Example 8.8. Find a series expansion centered at x = 0 for arctan(x), as well as its
interval of convergence.
Solution. While we could differentiate arctan(x) repeatedly and look for patterns in the
derivative values at x = 0 in an attempt to find the Maclaurin series for arctan(x) from
the definition, it turns out to be far easier to use a known series in an insightful way. In
Activity 8.29, we found that
1
1 + x 2 =
∞
k=0
(−1)
k x
2k
for −1 < x < 1. Recall that
d
dx
[arctan(x)] =
1
1 + x 2 ,
and therefore
1
1 + x 2 dx = arctan(x) + C.
It follows that we can integrate the series for
1
1 + x 2 term by term to obtain the power
series expansion for arctan(x). Doing so, we find that
arctan(x) =
∞
k=0
(−1)
k x
2k
dx
=
∞
k=0
(−1)
k x
2k dx
=
∞
k=0
(−1)
k x 2k+1
2k + 1
+ C.
The Power Series Differentiation and Integration Theorem tells us that this equality is
valid for at least −1 < x < 1.
To find the value of the constant C, we can use the fact that arctan(0) = 0. So
0 = arctan(0) =
∞
k=0
(−1)
k 0 2k+1
2k + 1
+ C = C,
and we must have C = 0. Therefore,
arctan(x) =
∞
k=0
(−1)
k x 2k+1
2k + 1
(8.27)