To find the Taylor Series expansion about origin of
$f(x)=\int_0^x \tan^{-1}t dt$
I have used $\tan^{-1}t=t-t^3/3+t^5/5-t^7/7+....$ and simply integrating I found f(x)= $\sum(-1)^{n-1}x^{2n}/(2n(2n-1))$ though I know the standard way to do is finding $f',f'',f'''$ and so on and their corresponding value in $0$ then using $f(x)=\sum (f^n(x)/n!) x^n$. Is the approach correct? Also using this result I have to find the sum of the series
$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+......$
I am not understanding how to do it? Any hint?
Hint: Rewrite this sum as $$\Bigl(1-\frac13+\frac15-\dotsm\Bigr)-\Bigl(\frac12-\frac14+\frac16-\dotsm\Bigr).$$