Parseval's Theorem for discrete series
You are right, but you have mistaken the DTFTs...

Considering the Parseval's relation for real discrete-time sequences to be:

$$ \sum_{n} x[n]y[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\omega})Y(e^{j\omega}) d\omega $$

We shall define $x_1[n]$, $x_2[n]$ and find their corresponding DTFT as:

$$ x_1[n] = \frac{\sin(\frac{1}{4} \pi n)}{2\pi n} \Leftrightarrow X_1(e^{j \omega}) = \begin{cases} \frac{1}{2} &amp;,&amp; \text{ for } |w| &lt; \frac{\pi}{4} \\ 0 &amp;,&amp; \text{otherwise} \end{cases} $$

$$ x_2[n] = \frac{\sin(\frac{1}{6} \pi n)}{5\pi n} \Leftrightarrow X_1(e^{j \omega}) = \begin{cases} \frac{1}{5} &amp;,&amp; \text{ for } |w| &lt; \frac{\pi}{6} \\ 0 &amp;,&amp; \text{otherwise} \end{cases} $$

Where we have used the DTFT pair;
$$ x[n] = A\frac{\sin(\omega_c  n)}{\pi n} \Leftrightarrow X(e^{j \omega}) = \begin{cases} A &amp;,&amp; \text{ for } |w| &lt; \omega_c \\ 0 &amp;,&amp; \text{otherwise} \end{cases} $$

Hence applying Parseval's theorem we see that:

$$ S = \sum_{n} \frac{\sin(\frac{1}{4} \pi n)}{2\pi n} \frac{\sin(\frac{1}{6} \pi n)}{5\pi n} = \frac{1}{2\pi} \int_{-\pi/6}^{\pi/6} \frac{1}{2}\frac{1}{5} d\omega = \frac{1}{2\pi} \frac{2\pi}{6} \frac{1}{10}  = \frac{1}{60} = 0.01667$$

To check this result you can use the following OCTAVE/matlab code for getting the sum in the time domain:

L = 128;    % select large enough L for better approximation
n = -L:L;
x =  sin(pi*n/4).*sin(pi*n/6)./(2*5*pi*pi.*n.*n);
x(L+1)= (1/4)*(1/6)/(2*5);
S = sum(x)
&gt;&gt; ans = 0.01667


which verifies that our calculation was correct.