Find the sum ofthe following series (with infinite terms):
$$2\sqrt{2},\frac{4}{\sqrt{3}},\frac{4\sqrt{2}}{3},.....$$

The series is an infinite Geometric progression with first term as $$2\sqrt{\ 2}$$ and common ratio $$\sqrt{\ \frac{2}{3}}$$
Now we know that sum of an infinite geometric progression = $$\frac{a}{1-r}$$ where r is the common ratio and a is first term
so we get sum =$$\frac{2\sqrt{\ 2}}{1-\sqrt{\ \frac{2}{3}}}$$
                      =$$=\ \frac{2\sqrt{\ 6}}{\sqrt{\ 3}-\sqrt{\ 2}}$$
Rationalizing the denominator
we get :$$=\frac{2\sqrt{\ 6}\left(\sqrt{\ 3}+\sqrt{\ 2}\right)}{3-2}$$
= $$2\sqrt{\ 6}\left(\sqrt{\ 3}+\sqrt{\ 2}\right)$$