Originally Posted by

**Krizalid** Your serie is equal to $\displaystyle \frac\pi4.$

Consider $\displaystyle \arctan x = \int_0^x {\frac{1}

{{1 + u^2 }}\,du} .$ So,

$\displaystyle \arctan x = \int_0^x {\frac{1}

{{1 + u^2 }}\,du} = \int_0^x {\left\{ {\sum\limits_{k = 0}^\infty {( - 1)^k u^{2k} } } \right\}\,du} .$

Finally

$\displaystyle \arctan x = \sum\limits_{k = 0}^\infty {( - 1)^k \left\{ {\int_0^x {u^{2k} \,du} } \right\}} = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k + 1} }}

{{2k + 1}}} ,\,\left| x \right| \le 1.$

Now set $\displaystyle x=1$ and we're done.