real analysis - Proving $limlimits_{x to 0} ({1 + sumlimits_{n = 1}^infty {{( - 1 )^n}{{( {frac{{sin nx}}{{nx}}})}^2}} } )= frac 12$

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!

Answer

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!

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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

$$1-1+1-1+1\cdots =\frac12$$