Inspired by Daniel Liu

Calculus Level 3

1 1 3 + 1 5 1 7 + 1 9 1 11 + = ? \large 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+ \ldots = \, ?

Give your answer to 3 decimal places.


The answer is 0.785.

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2 solutions

Rajen Kapur
Jun 28, 2015

Using ln 1 + x 1 x = 2 ( x + x 3 3 + x 5 5 + x 7 7 + . . . . ) \ln\dfrac{1+x}{1-x} = 2(x +\dfrac{x^3}{3} + \dfrac{x^5}{5} + \dfrac{x^7}{7}+ . . . . ) at x = i, the series equals 1 2 i ln 1 + i 1 i = 1 2 i ln ( i ) = 1 4 π 0.785 \dfrac{1}{2i} \ln\dfrac{1+i}{1-i} = \dfrac{1}{2i} \ln(i) = \dfrac{1}{4}\pi \approx 0.785 . Hint: i = e i π / 2 i = e^{i\pi/2} .

Or you can use the Maclaurin series for tan 1 ( x ) \tan^{-1} (x) .

Pi Han Goh - 5 years, 11 months ago

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Wolfram MathWorld says that the Maclaurin series for tan 1 ( x ) \tan^{-1} (x) can be used for 1 < x < 1 -1 < x < 1 . Can it be used for x = 1 x = 1 too?

Pranshu Gaba - 5 years, 10 months ago

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Yea, it should work for 1 -1 and 1 1 as well. See Gregory Series . Let me report the issue to them.

Pi Han Goh - 5 years, 10 months ago

I too used the same method to solve :)

Surya Prakash - 5 years, 11 months ago
Kaustav Sengupta
Jul 23, 2015

Leibniz's already prove that

Leibniz wrote a Solution for this problem before? You so sure?

Muhammad Arifur Rahman - 5 years, 10 months ago

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