Evaluation of $\lim_{n\rightarrow \infty} \int_{0}^{1}\frac{1}{1+x^2+x^4+.........+x^{2n}}dx\;,$ Where $n\in \mathbb{N.}$ Evaluation of $\displaystyle \lim_{n\rightarrow \infty} \int_{0}^{1}\frac{1}{1+x^2+x^4+.........+x^{2n}}dx\;,$ where $n\in \mathbb{N.}$
$\bf{My\; Solution:}$ First we will simplify $\displaystyle 1+x^2+x^4+.......+x^{2n} = \frac{1-x^{2n+2}}{1-x^2}.$
Given $0< x<1, \quad\displaystyle \lim_{n\rightarrow \infty}\frac{1-x^{2n+2}}{1-x^2} = \frac{1}{1-x^2}$.
$ \therefore \displaystyle \displaystyle \lim_{n\rightarrow \infty} \int_{0}^{1}\frac{1}{1+x^2+x^4+.........+x^{2n}}dx = \int_{0}^{1}(1-x^2)dx = \frac{2}{3}.$
Is my solution is correct? If not, then how can we solve the above question?
Thanks in advance.

We have that:
$$ I_n = \int_{0}^{1}\frac{1-x^{2}}{1-x^{2n+2}}\,dx =\frac{2}{3}+\int_{0}^{1}x^{2n+2}\frac{1-x^2}{1-x^{2n+2}}\,dx\tag{1}$$
where, over $[0,1]$:
$$0\leq \frac{1-x^2}{1-x^{2n+2}}\leq 1,\tag{2}$$
hence:
$$ \frac{2}{3}\leq I_n\leq \frac{2}{3}+\frac{1}{2n+3}\tag{3} $$
gives that the limit is $\frac{2}{3}$ as wanted.
