This problem requires the use of mathematical induction to prove the following formula for all natural numbers $n \geq 1$:

$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n \cdot (n+1)} = \frac{n}{n+1}.$

Step 1: Base Case

First, we verify the formula for the base case, $n = 1$:

$\frac{1}{1 \cdot 2} = \frac{1}{2}.$ And the right-hand side is: $\frac{1}{1+1} = \frac{1}{2}.$ Thus, the base case holds because both sides are equal.

Step 2: Inductive Step

Now, we assume that the formula holds for some arbitrary $n = k$, where $k \geq 1$. This is called the inductive hypothesis. We assume that:

$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k \cdot (k+1)} = \frac{k}{k+1}.$

We must now prove that the formula holds for $n = k + 1$, that is, we need to show:

$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k \cdot (k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2}.$

Step 3: Proof for $n = k + 1$

Using the inductive hypothesis, we can rewrite the left-hand side as:

$\frac{k}{k+1} + \frac{1}{(k+1)(k+2)}.$

Now, we combine the two terms on the left-hand side by finding a common denominator:

$\frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)}.$

Now, we can combine the numerators:

$\frac{k(k+2) + 1}{(k+1)(k+2)} = \frac{k^2 + 2k + 1}{(k+1)(k+2)}.$

Notice that the numerator can be factored as:

$k^2 + 2k + 1 = (k+1)^2.$

Thus, the left-hand side becomes:

$\frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}.$

This is exactly the right-hand side of the equation we wanted to prove for $n = k + 1$.

Step 4: Conclusion

Since both the base case and the inductive step have been proven, we can conclude by the principle of mathematical induction that the formula holds for all $n \geq 1$.

Thus, we have shown that:

$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n \cdot (n+1)} = \frac{n}{n+1}$ for all $n \geq 1$.

Would you like further clarification on any of the steps or additional examples of induction?