1 − 3 1 + 5 1 − 7 1 + 9 1 − 1 1 1 + … = ?
Give your answer to 3 decimal places.
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Or you can use the Maclaurin series for tan − 1 ( x ) .
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Wolfram MathWorld says that the Maclaurin series for tan − 1 ( x ) can be used for − 1 < x < 1 . Can it be used for x = 1 too?
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Yea, it should work for − 1 and 1 as well. See Gregory Series . Let me report the issue to them.
I too used the same method to solve :)
Leibniz's already prove that
Leibniz wrote a Solution for this problem before? You so sure?
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Using ln 1 − x 1 + x = 2 ( x + 3 x 3 + 5 x 5 + 7 x 7 + . . . . ) at x = i, the series equals 2 i 1 ln 1 − i 1 + i = 2 i 1 ln ( i ) = 4 1 π ≈ 0 . 7 8 5 . Hint: i = e i π / 2 .