Give a concrete sequence of rationals which converges to an irrational number and vice versa.

Give a concrete sequence of rationals which converges to an irrational number and vice versa....

My work

I could give a sequence of irrationals which converges to a rational number...

Let $r\in \mathbb Q,$ $$a_n=\frac {\sqrt 2} n+r$$ But I couldn't give a sequence of rationals which converges to an irrational.. Help me to work out.....

$a_n = \left( 1 +\dfrac{1}{n} \right)^n $ converges to $e$

How about $a_n$ = the decimal expansion of $\sqrt{2}$ up to the $n$-th place

A formal definition could be $$a_n = \lfloor 10^n\sqrt{2} \rfloor 10^{-n} $$

Take ratios of consecutive Fibonacci numbers: $\frac11,\frac21,\frac32,\frac53,\frac85,\dots$. It is well known that this converges to the golden ratio $\frac{1+\sqrt{5}}{2}$, which is irrational.

Consider the sequence $$0.1, 0.101, 0.101001, 0.1010010001, 0.101001000100001, 0.101001000100001000001,\dots.$$ This converges to $\alpha=0.101001000100001000001\cdots$. The decimal expansion of $\alpha$ is not ultimately periodic, so $\alpha$ is irrational.

Any Taylor series that converges to an irrational would work. E.g.

$$\frac{1}{0!}+\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots = e$$

Pick any irrational number, pick a function that calculates it given a rational input, then the taylor series of that function around that input will fulfill the requirements.