taylor expansion and limit of a series?? $f\left(x\right)=∫_0^x\tan ^{-1}tdt$
what is the taylor expansion about the origin of this function?
and how do i use this to get the limit of the series
$1-\frac{1}{2}-\frac {1}{3}+\frac {1}{4}+\frac {1}{5}-\frac{1}{6}-\frac {1}{7}.......$ 
i could get the limit by using concepts like rearranging the terms and got a different limit since it is a conditionally convergent series and can be made to converge to any real number.but how do i get the limit using this taylor expansion.please somebody help?
Answer to the second part is $\frac {π}{4}-\frac {\log 2}{2}$
 A: The Taylor expansion can be obtained by observing
$$ f(x) = \int_0^x\arctan(t)dt=x\arctan x - \frac{1}{2}\ln(1+x^2)$$
Using the Mercator series we have 
$$\ln(1+x^2) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^{2k}$$
and for the first summand there is the well-know expansion
$$x\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{2k+1}=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{2k+2}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1}x^{2k}$$
Therefore the sum of the two functions is
$$f(x)=\sum_{k=1}^\infty (-1)^{k+1}x^{2k}\left( \frac{1}{2k-1}- \frac{1}{2k}\right)= \sum_{k=1}^\infty\frac{(-1)^{k+1}}{2k(2k-1)}x^{2k} $$
$$=\frac{1}{2}x^2-\frac{1}{12}x^4+\frac{1}{30}x^6-\frac{1}{56}x^8+\frac{1}{90}x^{10} \cdots$$
Of course this result can also be obtained by termwise integrating the $\arctan$ series
$$f(x)=\int \left (\sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1}x^{2k-1}\right) dx
= \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1}\int x^{2k-1}dx
=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{2k(2k-1)}x^{2k}$$
Unfortunately I have no clue how to relate the Taylor series to your other part of the question.
A: Hint: $$f'\left(x\right)= \tan^{-1}x$$ for fundamental theorem of calculus then I believe that you can continue.
