I have to find out the Taylor series of $f(x) = \arctan(x)$ and prove that it converges to $f(x)$ for any $x \in (-1, 1) $.
So far I determined the Taylor series to $T_f(x) = \sum \limits_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{2k+1} $ . With the ratio test, I proved convergence for $x^2 \lt 1$ which is satisfied by the domain of definition of $x$.
But how do I continue proving that $T_f(x)$ converges to $\arctan(x)$ for $-1 \lt x \lt 1$ ? Splitting up the series in positive and negative parts didn't help me either.
EDIT: I just noticed the following:
$T_f = \sum \limits_{k=0}^{\infty} \frac{x^{4k+1}}{4k+1} - \sum \limits_{k=0}^{\infty} \frac{x^{4k+3}}{4k+3} $ is the series split up in a positive and negative one. If I derive both, I get $T'_f = \sum \limits_{k=1}^{\infty} x^{4k} - \sum \limits_{k=1}^{\infty} x^2 x^{4k}$ ; can I apply the formula for geometric series here?
EDIT 2 : If I apply $\sum \limits_{k=0}^{\infty} a_0 q^k $ with $x^{4k} = (x^4)^k $ thus $q=x^4$ I get $T'_f = \frac{1}{1-x^4} - \frac{x^2}{1-x^4} = \frac{1-x^2}{(1-x^2)(1+x^2)} = \frac{1}{1+x^2}$ which is the exact derivative of $\arctan(x)$. What do you think about that?
$$\limsup_{n\to +\infty}\sqrt[n]{\frac{1}{2n+1}}=1$$ proves that the radius of convergence of the power series is $1$.
After that, for any $x\in(-1,1)$ we may notice that: $$ \sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{2n+1} = \int_{0}^{x}\sum_{n\geq 0}(-1)^n t^{2n}\,dt = \int_{0}^{x}\frac{dt}{t^2+1}=\arctan(x)$$ and we're done.