Convergent sequence of irrational numbers that has a rational limit.

Question

Is it possible to have a convergent sequence whose terms are all irrational but whose limit is rational?

Answer 1

$$\bigg\{a_{n} = \frac{\sqrt{2}}{n} \bigg\}$$

Answer 2

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 $\lim\limits_{n\to\infty}\left(q + \frac 1 n \sqrt 2\right) = q$.

Answer 3

Here is another example: $$a_n=\frac{1}{n\pi}$$

Answer 4

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.

Answer 5

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:

$$1=\cfrac{\pi^2/9}{2-\pi^2/9+\cfrac{2\pi^2/9}{12-\pi^2/9+\cfrac{12\pi^2/9}{30-\pi^2/9+\cfrac{30\pi^2/9}{56-\pi^2/9+\cdots}}}}$$

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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:

$$1=\cfrac{e}{e+\frac{1}{e}-\cfrac{1}{e+\frac{1}{e}-\cfrac{1}{e+\frac{1}{e}-\cdots}}}$$

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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 $2$:

$$\prod_{k=1}^{\infty} \frac{\sqrt[2k-1]{e}}{\sqrt[2k]{e}} =\frac{e \sqrt[3]{e} \sqrt[5]{e} \sqrt[7]{e} \cdots}{\sqrt{e} \sqrt[4]{e} \sqrt[6]{e} \sqrt[8]{e} \cdots}=2$$