Infinitely

$S\quad =\quad \frac { 1 }{ \sqrt { 5 } } +{ Red }{ \frac { 1 }{ 5 } +\frac { 1 }{ 5\sqrt { 5 } } }+\dots \dots \dots ---\heartsuit \\ \frac { 1 }{ \sqrt { 5 } } S\quad =\quad \frac { 1 }{ 5 } +\frac { 1 }{ 5\sqrt { 5 } } +\dots \dots \dots ----\ddot { \smile } \\ Subtract:\quad \\ \ddot { \smile } -\quad \heartsuit \quad =\quad (1-\frac { 1 }{ \sqrt { 5 } } )S\quad =\quad \frac { 1 }{ 5 } \quad \\ \vdots \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \vdots \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \vdots \\ S\quad =\quad \frac { \sqrt { 5 } }{ \sqrt { 5 } (\sqrt { 5 } -1) } \Longrightarrow \quad 0.8...\\ Highest\quad integer\quad less\quad then\quad 0.8\quad is\quad 0$

Given that

$X = \dfrac{1}{\sqrt{5}}+\dfrac{1}{5} + \dfrac{1}{5 \sqrt{5}}+ \dfrac{1}{25}+ .........................$

Now it is easy to determine that it is a Geometric Progression. Now,

$a = \dfrac{1}{\sqrt{5}}$ and $r = \dfrac{1}{\sqrt{5}}$

Now we know that, in a G.P, sum of infinite terms is given by

$\large{S_{\infty}}= \dfrac{a}{1-r}.$

Therefore,

${S_{\infty}} = \dfrac{\dfrac{1}{\sqrt{5}}}{1 - \dfrac{1}{\sqrt{5}}} = \dfrac{1}{\sqrt{5}-1} = \dfrac{\sqrt{5}-1}{4}$

We know that $\sqrt{5} = 2.236$ (approx)

Now,

$X = \dfrac{2.236+1}{4} = \dfrac{3.236}{4} = 0.8.$

Therefore,

$\left \lfloor X \right \rfloor = \left \lfloor 0.8 \right \rfloor = \boxed{\boxed{0}}$