To find $P(X \geq 3)$, we first determine the constant $k$ using the total probability for $X$.

The probability $P(X = x)$ is given by:

$P(X= x) = k(x+1) \cdot 3^{-x}, x= 0,1,2,3,\dots$

The total probability must equal 1:

$s = \sum_{x=0}^{\infty} k(x+1) \cdot 3^{-x}$

Calculating that series:

$s = k3^0 + 2\frac{k}{3} + 3\frac{k}{3^2} + \dots$

Therefore, dividing the series by 3:

$\frac{s}{3} = \frac{k}{3} + 2\frac{k}{3^2} + \dots$

Subtracting these:

$s - \frac{s}{3} = k + \frac{k}{3} + \frac{k}{3^2} + \dots$

The resulting series is a geometric series:

$2\frac{s}{3} = k(1 + \frac{1}{3} + \frac{1}{3^2} + \dots )$

The sum of the infinite geometric series is:

$2\frac{s}{3} = k \cdot \frac{1}{1 - \frac{1}{3}} = \frac{3k}{2}$

Equating:

$s = \frac{9k}{4} = 1$

Thus, solving for $k$:

$k = \frac{4}{9}$

Next, compute $P(X \geq 3)$:

$P(X \geq 3) = 1 - (P(X=0) + P(X=1) + P(X=2))$

Calculating these:

$P(X=0) = k = \frac{4}{9}$

$P(X=1) = 2\frac{k}{3} = \frac{8}{27}$

$P(X=2) = 3\frac{k}{9} = \frac{4}{27}$

Adding these probabilities:

$P(X=0) + P(X=1) + P(X=2) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{8}{9}$

Finally, calculate $P(X \geq 3)$:

$P(X \geq 3) = 1 - \frac{8}{9} = \frac{1}{9}$