Considering the power series $\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n+1)(2n-1)}$. Consider the power series $\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n+1)(2n-1)}$. Find a closed form expression for all x which converge and hence evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)(2n-1)}$.
Attempt at the solution: The radius of convergence is 1. We can rewrite the summands by:
\begin{eqnarray}
\sum_{n=1}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)(2n-1)} &=& \frac{1}{2}\Big[(x^3-\frac{x^3}{3})  - (\frac{x^5}{3} - \frac{x^5}{5}) + (\frac{x^7}{5} - \frac{x^7}{7}) + \dots \Big]\\
&=& \frac{1}{2}\Big[(x^3  -\frac{x^5}{3}+ \frac{x^7}{5} + \dots ) + (\frac{-x^3}{3} + \frac{x^5}{5}-\frac{x^7}{7} +...)\Big]\\
&=& \frac{1}{2}\Big[x^2\int\frac{1}{1+x^2}dx + \int\frac{1}{1+x^2}dx-x\Big]\\
&=& \frac{x^2}{2}\arctan(x)+\frac{1}{2}\arctan(x) -\frac{x}{2}
\end{eqnarray}
Substituting $x=1$ then gives $\frac{\pi}{4}-\frac{1}{2}$ .

The issue I have is two fold. Firstly, when dealing with evaluations at the boundary, term by term differentiation may not be valid. In particular, we used the fact that $\arctan(x) = x-\frac{x^3}{3} + ...$ by integrating power series for $\frac{1}{1+x^2}$, valid for |x|<1. This means that the arctan formula can only be guaranteed to hold within the interior (-1,1). What are the conditions needed to talk about power series validity at boundary points?

(Abelian/Tauberian theorems came to mind at first, but the conditions in this problem weren't strong enough. Alternatively, I noted that uniform convergence of the terms meant that the limit function of $x-\frac{x^3}{3} + ...$ had to be continuous. So $\arctan(1) = \frac{\pi}{4}$ by continuous extension. Do correct me if I'm wrong.
The other issue that I have not been able to justify is that of conditional convergence. Clearly, the arctan series is conditionally convergent at $x=1$. How do we justify the rearrangments carried out above then?
 A: First, by using the standard power series,
$$
\sum_{n=1}^\infty(-1)^{n-1}\frac{t^{2n-1}}{2n-1}=\arctan t, \quad |t|<1,\tag1
$$ we multiply $(1)$ by $t$ then we are alowed to integrate termwise:
$$
\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n-1)(2n+1)}=\int_0^x t\arctan t\: dt, \quad |x|<1,\tag2
$$ then integrating by parts on the right hand side, gives
$$
\int_0^x t\arctan t\: dt=\frac12\left(1+x^2\right)\arctan x-\frac{x}2, \tag3
$$ thus

$$
\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n-1)(2n+1)}=\frac12\left(1+x^2\right)\arctan x-\frac{x}2, \quad |x|<1.\tag4
$$ 

Second, noticing the following absolute convergence,
$$
\sum_{n=1}^\infty\left|(-1)^{n-1}\frac{1}{(2n-1)(2n+1)}\right|<\infty,
$$ one may use Abel's theorem in $(4)$ with $x \to 1^-$ to obtain

$$
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)(2n+1)}=\frac{\pi}4-\frac12 \tag6
$$

as announced.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&\color{#f00}{\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{x^{2n + 1} \over \pars{2n + 1}\pars{2n - 1}}} =
\half\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{x^{2n + 1} \over 2n - 1} - \half\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{x^{2n + 1} \over 2n + 1}
\\[4mm] = &\
\half\,x^{3} + \half\,x^{2}\sum_{n = 1}^{\infty}
\pars{-1}^{n}\,{x^{2n + 1} \over 2n + 1} - \half\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{x^{2n + 1} \over 2n + 1}
\\[4mm] = &\
\half\,x^{3} + \half\,\pars{x^{2} + 1}x\
\underbrace{\sum_{n = 1}^{\infty}
\pars{-1}^{n}\,{x^{2n} \over 2n + 1}}_{\ds{\equiv\ \,\mathcal{J}}}\tag{1}
\end{align}

\begin{align}
\,\mathcal{J} & =
\sum_{n = 1}^{\infty}
\pars{-1}^{n}\,{x^{2n} \over 2n + 1} =
\sum_{n = 1}^{\infty}\pars{-1}^{n}\,x^{2n}\int_{0}^{1}y^{2n}\,\dd y =
\int_{0}^{1}\sum_{n = 1}^{\infty}\pars{-x^{2}y^{2}}^{n}\,\dd y
\\[4mm] &\ =\
\overbrace{\int_{0}^{1}{-x^{2}y^{2} \over 1 + x^{2}y^{2}}\,\dd y}
^{\ds{-1 + {\arctan\pars{x} \over x}}}
\end{align}

$$
\color{#f00}{\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{x^{2n + 1} \over \pars{2n + 1}\pars{2n - 1}}} =
\color{#f00}{\half\bracks{\pars{x^{2} + 1}\arctan\pars{x} - x}} 
$$


When $\ds{x \to 1^{-}}$, $\ds{\sum\cdots \to {\pi \over 4} - \half}$.

