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Is there a non recursive, explicit sequence of rational numbers that has $\sqrt{2}$ as a limit?

I know of rational sequences such as $x_{n+1}=(x_n+2/x_{n})/2$ and $q_n=[10^n\sqrt{2}]/10^n$ that have $\sqrt{2}$ as limit (as mentioned in this other post): tho, the first is defined by recursion, the latter already has $\sqrt{2}$ in its explicit form; so I fell like they're missing the point of my question.

I wonder if is there a rational sequence $a_n$ that has $\sqrt{2}$ as limit whose terms can be written explicitly in terms of elementary functions that take $n$ as the argument. I've tried to find one but nothing seems to make sense. If there is not such a sequence, could you explain why?

Is there a non-recursive, explicit sequence of rational numbers that has $\sqrt{2}$ as a limit?

I know of rational sequences such as $x_{n+1}=(x_n+2/x_{n})/2$ and $q_n=[10^n\sqrt{2}]/10^n$ that have $\sqrt{2}$ as limit (as mentioned in this other post), though the first is defined by recursion, and the latter already has $\sqrt{2}$ in its explicit form, so I feel like they're missing the point of my question.

I wonder if is there a rational sequence $a_n$ that has $\sqrt{2}$ as limit whose terms can be written explicitly in terms of elementary functions that take $n$ as the argument. I've tried to find one but nothing seems to make sense. If there is not such a sequence, could you explain why?

Is there a non recursive, explicit sequence of rational numbers that has $\sqrt{2}$ as a limit?

I know of rational sequences such as $x_{n+1}=(x_n+2/x_{n})/2$ and $q_n=[10^n\sqrt{2}]/10^n$ that have $\sqrt{2}$ as limit (as mentioned in this other post): tho, the first is defined by recursion, the latter already has $\sqrt{2}$ in its explicit form; so I fell like they're missing the point of my question.

I wonder if is there a rational sequence $a_n$ that has $\sqrt{2}$ as limit whose terms can be written explicitly in terms of elementary functions that take $n$ as the argument. I've tried to find one but anything seems to make sense. If there is not such a sequence, could you explain why?

Is there a non recursive, explicit sequence of rational numbers that has $\sqrt{2}$ as a limit?

I know of rational sequences such as $x_{n+1}=(x_n+2/x_{n})/2$ and $q_n=[10^n\sqrt{2}]/10^n$ that have $\sqrt{2}$ as limit (as mentioned in this other post): tho, the first is defined by recursion, the latter already has $\sqrt{2}$ in its explicit form; so I fell like they're missing the point of my question.

I wonder if is there a rational sequence $a_n$ that has $\sqrt{2}$ as limit whose terms can be written explicitly in terms of elementary functions that take $n$ as the argument. I've tried to find one but nothing seems to make sense. If there is not such a sequence, could you explain why?