Series Everywhere

The value of $N$ could be simplified as

$\large N =\huge \frac { 6 \times (\sum _{n=1}^{\infty }\frac{1}{n^2})^{2}}{ \sum_{n=1}^{\infty \:}\frac{1}{n^4} }$

$\large N = \huge \frac { 6 \times (\frac {\pi^{2}}{6})^{2} }{ \frac { \pi^{4}}{90}}$

Then, by doing the necessary simplifications and cancellations, we get

$\large N = 15$

Then, substituting the value of $N$ , we get

$\large E = \huge { \frac{15}{2+\frac{15}{2+\frac{15}{2+\frac{15}{2+\frac{15}{2+...}}}}}}$

Since this is a continued fraction, we can simplify this and get

$\large E = \huge { \frac {15}{2 + E} }$

$\large E(2+E) = 15$

$\large E^{2} + 2E - 15 = 0$

$\large (E+5)(E-3) = 0$

since $E$ has to be positive, we get

$\boxed { \large E = 3 }$ ,

and thus

$\boxed { \large E^{2} = 9 }$ .