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$f(x)=\int_0^xtan^{-1}tdt$

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 $\frac {\pi}{4}-\frac {log2}{2}$

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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.

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Hint: $$f'(x)= tan^{-1}x$$ for fundamental theorem of calculus then I believe that you can continue.

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  • $\begingroup$ No i cant continue.please give a bit more.. $\endgroup$ – user311790 Feb 16 '16 at 12:16
  • $\begingroup$ Do you know the Taylor expansion of $\arctan x$? $\endgroup$ – Domenico Vuono Feb 16 '16 at 12:20
  • $\begingroup$ $x-x^3/3+x^5/5-x^7/7.........$ $\endgroup$ – Upstart Feb 16 '16 at 12:33
  • $\begingroup$ Yea now you can use for the other derivatives you can use the Taylor expansion of arctan x but you have focus on coefficients of $x, x^3.... $\endgroup$ – Domenico Vuono Feb 16 '16 at 12:40
  • $\begingroup$ ???? i don't understand what are you trying to hint at? $\endgroup$ – Upstart Feb 16 '16 at 12:42

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