An Intriguing Result

Calculus Level pending

$\large 4 \times \sum_{n=1}^\infty \dfrac{-1 + 2(n \bmod 2)}{2n-1} =\, ?$

2\sqrt{2}

\infty

\frac{2}{\sqrt{0.5}}

e

\pi

1 solution

Xiangchen Kong
Feb 2, 2016

If we put n=1,2,3,4…,n into the sum equation above

we can clearly see that it is equivalent to 4 * ( 1-1/3+1/5-1/7…+（-1）^n *(1/2n+1) )

which can fit in to the Maclaurin's series as below

tan-1(x)= x-x^3/3+x^5/5-…+(-1)^r* ((x^2r+1)/(2r+1)) (-1<=x<=1)

where x=1

tan-1(x) is clearly π/4 (45˚)

therefore 4 times the result above(π/4) is π.