Show that the series $\sum_{k=0} ^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$ Show that the series $\sum_{k=0} ^\infty (-1)^k \dfrac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$
I tried using the ratio test but I got that it equals 1, so it is inconclusive. How can I show that it converges, and that it converges to arctan x? I think I'm trying to show that the Legrange error term is 0 but I'm not sure how to proceed with it - the nth derivatives of arctan x keep changing so...

Modifying Jack D'Aurizio's answer,
since
$\sum_{k=0}^n x^k = \dfrac{1-x^{n+1}}{1-x}$,
putting $-x^2$ for $x$ we get
$\sum_{k=0}^n (-1)^kx^{2k} 
= \dfrac{1-(-1)^{n+1}x^{2n+2}}{1+x^2}
= \dfrac1{1+x^2}+\dfrac{(-1)^{n}x^{2n+2}}{1+x^2}
$
or
$\dfrac1{1+x^2}
=\sum_{k=0}^n (-1)^kx^{2k} -\dfrac{(-1)^{n}x^{2n+2}}{1+x^2}
$.
Integrating from $0$ to $y$,
$\begin{array}\\
\arctan(y)
&=\int_0^y \dfrac{dx}{1+x^2}\\
&=\int_0^y \left(\sum_{k=0}^n (-1)^{k-1}x^{2k}-(-1)^{n}\dfrac{x^{2n+2}}{1+x^2}\right)dx\\
&=\sum_{k=0}^n (-1)^{k-1}\int_0^y x^{2k}dx-(-1)^{n}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\
&=\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}-(-1)^{n}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\
\end{array}
$
so
$\arctan(y)-\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}
=(-1)^{n+1}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}
$
so that
$\begin{array}\\
|\arctan(y)-\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}|
&\le\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\
&\le\int_0^y x^{2n+2}dx\\
&=\dfrac{x^{2n+3}}{2n+3}\\
&\to 0
\qquad\text{as } n \to \infty\\
\end{array}
$
