Can we say that $\sqrt{2}= \cfrac{2}{\cfrac{2}{\cfrac{2}{\cfrac{2}{\vdots}}}}$?
We have $$\phi=1+\cfrac{1}{\phi}=1+\cfrac{1}{1+\cfrac{1}{\phi}}=1+\cfrac{1}{1+\cfrac{1}{1+\frac{1}{\phi}}}=\cdots$$
(with $\phi$ being the Golden Ratio)
Which gives us the confirmed infinite fraction $$\phi=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}$$
We also have $$\sqrt{2}=\cfrac{2}{\sqrt{2}}=\cfrac{2}{\cfrac{2}{\sqrt{2}}}=\cfrac{2}{\cfrac{2}{\cfrac{2}{\sqrt{2}}}}=\cdots$$
So by analogy we can deduce that $$\sqrt{2}=\cfrac{2}{\cfrac{2}{\cfrac{2}{\cfrac{2}{\vdots}}}}$$
The sequence $(a_n)_{n\in\mathbb Z^+}$ such that $a_1=\sqrt{2}, a_{n+1}=\frac{2}{a_n}$ gives $\lim_{n\to +\infty}a_n=\sqrt{2}$, so indeed the representation should be correct.
Everywhere on the Internet that I see a continued fraction of $\sqrt{2}$, it is $$\sqrt{2}=1+(\sqrt{2}-1)=1+\cfrac{1}{1+\sqrt{2}}=1+\cfrac{1}{2+\cfrac{1}{1+\sqrt{2}}}=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{1+\sqrt{2}}}}=\cdots$$
This gives $$\sqrt{2}=1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{\ddots}}}}}$$
Why haven't I seen the representation $\sqrt{2}=2/(2/(2/(2/\dots)))$ mentioned, and see the above used instead? Is something wrong about my representation?
I would say it isn't mentioned because it is not useful: you cannot approximate $\sqrt{2}$ using the representation, unlike you can using the above one (which you can do by removing $\frac{1}{1+\sqrt{2}}$ in any member).
The problem is that if you take $a_0$ non-zero and you define $a_n = 2/a_{n-1}$, you have the relation $a_n = a_{n-2}$ (check that) so that the sequence converges if and only if $a_n = a_{n-1}$, which implies $a_0 = \pm \sqrt{2}$. So we cannot "deduce by analogy" because there is no analogy to make. The difference with the continued fractions is that in this case, the recurrence relation building the pattern with the $\cdots$ dots gives rise to a sequence which converges.
Hope that helps,