real analysis - On the series $sum limits_{n=1}^{infty} left ( frac{1}{n} - frac{1}{2n-1} - frac{1}{2n+1} right )$

I managed to prove through complex analysis that

$$\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right ) = 1 -2 \log 2$$

However, I'm having a difficult time proving this result with real analysis methods. Partial summation of the series gets nasty pretty quickly since it involves harmonic numbers. Another interesting fact to note about this series is that this part

$$\sum_{n=1}^{\infty} \left ( \frac{1}{2n-1} + \frac{1}{2n+1} \right )$$

diverges. This came as a complete surprise to me since I expected this to telescope. What remains now to prove the series I want is by using generating function.

\begin{align*}
\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )x^n &=\sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{2n-1} - \sum_{n=1}^{\infty} \frac{x^n}{2n+1} \\
&= -\log \left ( 1-x \right ) - \sqrt{x} \;\mathrm{arctanh} \sqrt{x} - \frac{{\mathrm {arctanh} }\sqrt{x} - \sqrt{x}}{\sqrt{x}}
\end{align*}

I cannot , however, evaluate the limit as $x \rightarrow 1^-$ of the last expression. Can someone finish this up?

Of course alternatives are welcome.

Answer

A very simple way:
$$ \begin{eqnarray*}\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{2n-1}-\frac{1}{2n+1}\right)&=&\sum_{n\geq 1}\int_{0}^{1}\left( 2x^{2n-1}-x^{2n-2}-x^{2n}\right)\,dx\\&=&\int_{0}^{1}\sum_{n\geq 1}x^{2n-2}(2x-1-x^2)\,dx\\&=&\int_{0}^{1}\frac{2x-1-x^2}{1-x^2}\,dx\\&=&\int_{0}^{1}\frac{x-1}{x+1}\,dx=\left[x-2\log(1+x)\right]_{0}^{1}=\color{red}{1-2\log 2}.\end{eqnarray*} $$