Question

\[=\sum\limits_{r=1}^\infty\frac{(-1)^r}{1-2r}\]
\[=1-\frac13+\frac15-\frac17+\dots\]

Answer 1

\[\large=\frac\pi4\]

Answer 2

@experimentX

Answer 3

http://tinyurl.com/d5x9bk9

Answer 4

\[ \int_0^1{1 \over 1+x^2}dx = \arctan(1) = {\pi \over 4} \]

Answer 5

http://www.wolframalpha.com/input/?i=expand+1%2F%281%2Bx^2%29

Answer 6

\[ f(u) = \int_0^u {1\over 1+x^2}dx = \int_0^u(1-x^2 + x^4 - x^5 + ...)dx \\\]
put 'u=1'

Answer 7

hmm