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\[=\sum\limits_{r=1}^\infty\frac{(1)^r}{12r}\]
\[=1\frac13+\frac15\frac17+\dots\]
 one year ago
 one year ago
\[=\sum\limits_{r=1}^\infty\frac{(1)^r}{12r}\] \[=1\frac13+\frac15\frac17+\dots\]
 one year ago
 one year ago

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UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[\large=\frac\pi4\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ \int_0^1{1 \over 1+x^2}dx = \arctan(1) = {\pi \over 4} \]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=expand+1%2F%281%2Bx^2%29
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ f(u) = \int_0^u {1\over 1+x^2}dx = \int_0^u(1x^2 + x^4  x^5 + ...)dx \\\] put 'u=1'
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
dw:1356251299233:dw
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
I guess he was trying to say that
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
dw:1356251680713:dw
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
So,dw:1356251895939:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yep!! that's correct ...
 one year ago
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