Convergent sequence of irrational numbers that has a rational limit. Is it possible to have a convergent sequence whose terms are all irrational but whose limit is rational? 
 A: $$\bigg\{a_{n} =  \frac{\sqrt{2}}{n} \bigg\}$$
A: I will give a more interesting answer (I think OP wants something like that):
$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}=2$$
Generally
$$\sqrt{a+b\sqrt{a+b\sqrt{a+b\sqrt{a+\cdots}}}}=\frac{1}{2}(b+\sqrt{b^2+4a})$$
It's not hard to find such numbers that $\sqrt{b^2+4a}$ is rational.
Also:
$$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^...}}=2$$
Also, using Euler's continued fraction theorem we can have something like this:
$$1=\cfrac{\pi^2/9}{2-\pi^2/9+\cfrac{2\pi^2/9}{12-\pi^2/9+\cfrac{12\pi^2/9}{30-\pi^2/9+\cfrac{30\pi^2/9}{56-\pi^2/9+\cdots}}}}$$

Actually, I can do even better. Let $\phi$ be the golden ratio, then we have:
$$1=\frac{1}{\phi^2}+\frac{1}{\phi^3}+\frac{1}{\phi^4}+\frac{1}{\phi^5}+\cdots=\sum^{\infty}_{k=2}\frac{1}{\phi^k}$$
But we don't want $e$ to feel left out, so here is another one:
$$1=\cfrac{e}{e+\frac{1}{e}-\cfrac{1}{e+\frac{1}{e}-\cfrac{1}{e+\frac{1}{e}-\cdots}}}$$

Another good one. Using the following:
$$2=e^{\ln 2}=e^{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots}$$
We obtain an infinite product, converging to $2$:
$$\prod_{k=1}^{\infty} \frac{\sqrt[2k-1]{e}}{\sqrt[2k]{e}} =\frac{e \sqrt[3]{e} \sqrt[5]{e} \sqrt[7]{e} \cdots}{\sqrt{e} \sqrt[4]{e} \sqrt[6]{e} \sqrt[8]{e} \cdots}=2$$
A: If $q$ is any rational number at all and $n$ is a positive integer then $q+\frac 1 n \sqrt 2$ is irrational (it's a simple algebra exercise to prove that), and $\lim\limits_{n\to\infty}\left(q + \frac 1 n \sqrt 2\right) = q$.
A: Here is another example: $$a_n=\frac{1}{n\pi}$$ 
A: Let $q$ be a rational number, $\alpha$ be an irrational number, and $a_n$ be a sequence of integers where $a_n \to \infty$. Then the sequence $$b_n = q + \frac {\alpha} {a_n}$$ satisfies the property you asked.
