Intersection of a nested interval of $A_n=\left[3-{\frac{1}{\sqrt{n}},3+\frac{1}{3^n}}\right]$ 
$A_n=\left[3-{\frac{1}{\sqrt{n}},3+\frac{1}{3^n}}\right]$
  What is $\bigcap_{n=1}^{\infty}A_n$

Since every set becomes a subset of the next set, is it correct to say that the intersection of all sets $n$ is just:
$\left[2,\frac{10}{3}\right]$ since this is the only set in all of the sets?

$$3+\frac{1}{3^n}-(3+\frac{1}{3^{n-1}}) = \frac{-2}{3^{n}} < 0$$ for all $ n \ge 2$ (1)
$$3-{\frac{1}{\sqrt{n}}}-(3-{\frac{1}{\sqrt{n-1}}}) = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n-1}\sqrt{n}} > 0 $$ for all $ n \ge 2$(2)
$$\lim_{x\to \infty}3+\frac{1}{3^n} = 3$$
$$\lim_{x\to \infty}3-{\frac{1}{\sqrt{n}}} = 3$$
Therefore By (1), $$3+\frac{1}{3^n} \ge 3$$
Therefore By (2), $$3\ge 3-{\frac{1}{\sqrt{n}}}$$
So the intersection set is $\{3\}$
