Sum the cosine limit

The limit is $1^{\infty}$ form , hence

$f(x)=e^{x^2\displaystyle\lim_{n\to\infty}\frac{ \left(\cos \left(\dfrac{x}{\sqrt{n}}\right)-1\right)}{\frac{x^2}{n}}}=e^{-x^2/2}$

Alternatively:- $\small{\begin{aligned}f(x)=e^{\displaystyle\lim_{n \rightarrow \infty} n\ln \left(\cos \left(\dfrac{x}{\sqrt{n}}\right)\right)}=&e^{\displaystyle\lim_{n \rightarrow \infty} \frac{\ln \left(\cos \left(\dfrac{x}{\sqrt{n}}\right)\right)}{\frac 1n}}\\=& e^{\displaystyle\lim_{n \rightarrow \infty} \frac{\ln \left(\color{#D61F06}{\cos \left(\dfrac{x}{\sqrt{n}}\right)-1}+1\right)}{\color{#D61F06}{\left(\cos \dfrac{x}{\sqrt{n}}-1\right)}\frac 1n}\times\left(\cos \dfrac{x}{\sqrt{n}}-1\right)}\\=&e^{x^2 \displaystyle\lim_{n\to\infty}\dfrac{\left(\cos \dfrac{x}{\sqrt{n}}-1\right)}{\frac{x^2}{n}}}=e^{-\frac{x^2}{2}}\end{aligned}}$

Apply Lhopital to see $f(x)=e^{-x^2/2}$ .

Hence $g(x)=\dfrac{1}{x^2}$

$\therefore \sum_{n=1}^{\infty}g(n)=\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\zeta (2)=\dfrac{\pi^2}{6}\approx \boxed{1.6449}$