Which term of the series $\frac{\sqrt[]{5}}{3},\, \frac{\sqrt[]{5}}{4},\frac{1}{\sqrt[]{5}},\, ...$ is $\frac{\sqrt{5}}{13}$ ?

Step 1: Rewrite all terms with same numerator
$\dfrac{\sqrt{5}}{3},\ \dfrac{\sqrt{5}}{4},\ \dfrac{\sqrt{5}}{5},\ \ldots$

(Since $\dfrac{1}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}$)

Step 2: Identify the AP in denominators
Denominators: $3,\ 4,\ 5,\ \ldots$
First term $a = 3$
Common difference $d = 1$

Step 3: Set up the equation
The $n^{th}$ term of the series $= \dfrac{\sqrt{5}}{a + (n-1)d}$

Given $n^{th}$ term $= \dfrac{\sqrt{5}}{13}$

So denominator of $n^{th}$ term $= 13$

Step 4: Solve for $n$
$a + (n-1)d = 13$
$3 + (n-1) \times 1 = 13$
$n - 1 = 10$
$n = 11$

Answer: $\dfrac{\sqrt{5}}{13}$ is the $\boxed{11^{th}}$ term of the series.