Can i do this to an infinite series? $$let , Y=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}}}$$
$$Then , Y= \sqrt{2Y}$$
$$Y^2 = 2Y$$
$$Y^2 - 2Y = 0$$
$$Y = 0 , Y =2$$
Now y can't be zero so ,
$$Y = 2$$
Is This Correct?

If you assume the number exists and has a value then that is true.
But it could be that the number doesn't exist.
Consider $1 + 2 + 4 + 8 + 16+..... = M$ then $2M = 2+4 + 8 + 32 + ....$ and $2M + 1 = 1 + 2 + 4+ 8+16 + 32 + .... = M$ so $2M + 1 = M$ and $M = -1$.
That's obviously not true.
ANd this fails because $\lim\limits_{n\to \infty}\sum\limits_{k=0}^n 2^k=\infty$.
However  If $a_0 =\sqrt{2}$ and $a_1= \sqrt {2\sqrt 2}$ and $a_{k+1} =\sqrt {2a_k}$ then IF $\{a_n\}$ converges so $Y = \lim\limits_{n\to \infty} a_n\ne \infty$, we would be okay.  And we could say $Y=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}}}$ and $Y= 2$ by your argument.
We can prove $\lim\limits_{n\to \infty} a_n\ne \infty$ by noting that $a_0=\sqrt 2 < 2$ and if $a_k <2$ we would have $a_{k+1} =\sqrt {2a_k}< \sqrt {2*2} = 2$. so by induction all $a_n < 2$ so $\lim\limits_{n\to \infty} a_n\ne \infty$.
Furthermore $a_{k+1} = \sqrt{2a_k} > \sqrt{a_k*a_k} = a_k$.
So we have $\sqrt{2} =a_1 < a_2 < a_3 < ....... < 2$.  So $\lim\limits_{n\to \infty} a_n = Y$ exists.
