Proving $\lim\limits_{x \to 0} ({1 + \sum\limits_{n = 1}^\infty {{( - 1 )^n}{{( {\frac{{\sin nx}}{{nx}}})}^2}} } )= \frac 12$ 
Could you show me $$\mathop {\lim }\limits_{x \to 0} \left( {1 + \sum\limits_{n = 1}^\infty  {{{\left( { - 1} \right)}^n}{{\left( {\frac{{\sin nx}}{{nx}}} \right)}^2}} } \right) = \frac{1}{2}.\tag{1}$$

These day, I want to write a article about the sums of divergent series. In Hardy's book Divergent Series, he show me a method proposed by Riemann with
$$\mathop {\lim }\limits_{x \to 0} \left( {1 + \sum\limits_{n = 1}^\infty  {{{\left( { - 1} \right)}^n}{{\left( {\frac{{\sin nx}}{{nx}}} \right)}^2}} } \right) ,$$ from which we can obtain $$1-1+1-1+\cdots=\frac12.$$
I hope someone tell me how to prove (1),Thanks!
 A: We begin with the function $f(x)$ as represented by the series
$$f(x)=\sum_{n=1}^\infty (-1)^n\frac{\sin^2 nx}{n^2x^2} \tag 1$$
Using the identity 
$$\sin^2nx=\frac{1-\cos 2nx}{2}$$
in $(1)$ yields
$$\begin{align}
f(x)&=\frac12 \sum_{n=1}^\infty (-1)^n\frac{1-\cos  2nx}{n^2x^2} \\\\
&=\frac12 \sum_{n=1}^\infty \frac{(-1)^n}{n^2x^2}-\frac12\text{Re}\left(\sum_{n=1}^\infty (-1)^n\frac{e^{i2nx}}{n^2x^2}\right)\\\\
&=-\frac{\pi^2}{24x^2}-\frac{1}{2x^2}\text{Re}\left(\text{Li}_2\left(-e^{i2x}\right)\right) \tag 2
\end{align}$$
We expand $\text{Li}_2\left(-e^{i2x}\right)$ in a series around $x=0$ to find 
$$\text{Li}_2\left(-e^{i2x}\right)=-\frac{\pi^2}{12}-i2\log (2)x+ x^2+O(x^3) \tag 3$$
Then, upon substituting $(3)$ into $(2)$ we have
$$f(x)=-\frac12 +O(x)$$
Finally, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\left(1+\sum_{n=1}^\infty (-1)^n\frac{\sin^2 nx}{n^2x^2} \right)=\frac12}$$
as was to be shown!

NOTE:
If we inappropriately interchange the limit with the series and use $\lim_{x\to 0}\frac{\sin^2 nx}{n^2x^2}=1$, we obtain the absurd conclusion that 
$$1-1+1-1+1\cdots =\frac12$$
A: Taking partial sums of the series and evaluating the limits, you get the following identity
$$\lim_{x \to 0} \sum\limits_{n = 1}^\infty  (-1)^n{\left(\frac{\sin nx}{nx} \right)^2} = 1+\frac{-\pi^2-6 \text{Li}_2(-e^{-2 i x})-6 \text{Li}_2(-e^{2 i x})}{24 x^2}$$
Now let's take the limit
$$\lim_{x \to 0} \Bigg(1+\frac{-\pi^2-6 \text{Li}_2(-e^{-2 i x})-6 \text{Li}_2(-e^{2 i x})}{24 x^2}\Bigg)$$
$$ = 1+ \frac{1}{24}\lim_{x \to 0}\frac{-\pi^2-6 \text{Li}_2(-e^{-2 i x})-6 \text{Li}_2(-e^{2 i x})}{x^2}$$
$$ = 1+ \frac{1}{48}\lim_{x \to 0}\left[-\frac{24e^{-2ix}}{1+e^{-2ix}}-\frac{24e^{2ix}}{1+e^{2ix}}\right]$$
$$ = 1+ \frac{1}{48}\left[-\frac{24}{2}-\frac{24}{2}\right]$$
$$ = 1- \frac{24}{48} = 1-\frac{1}{2} = \color{red}{\frac{1}{2}}$$
