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**Unformatted text preview: **f convergence of this series is 1. At x = 1, the series becomes
∞ (−1)n
n=0 1
2n + 1 which converges by the Alternating Series Test.
When x = −1, the series is
∞ ∞ 2n+1
n (−1) (−1) 2n + 1 n=0 (−1)3n+1 =
n=0 1
.
2n + 1 Since
1
if n is odd
−1 if n is even (−1)3n+1 =
the last series is the same as ∞ (−1)n+1
n=0 1
2n + 1 which also converges by the Alternating Series Test.
Therefore, the Continuity Theorem for Power Series shows that
∞ (−1)n arctan(x) =
n=0 x2n+1
2n + 1 for all x ∈ [−1, 1]
This last statement has an interesting application. We get that
π
4 = arctan(1)
∞ (−1)n =
n=0 DE Math 128 339 1
2n + 1
(B. Forrest)2 4.5. Introduction to Taylor Series CHAPTER 4. Sequences and Series Multiplying both sides of this equation by 4 gives
1111
+ − + − ···)
3579
4444
= 4 − + − + − ···
3579 π = 4(1 − The series representation for arctan(x) given above is called Gregory’s series after the
Scottish mathematician of the same name. The famous series expansion for π which we
derived from Gregory’s series is called Leibniz’s formula...

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