To solve the limit of the given sequence:

$S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)}$

We need to simplify the general term $\frac{1}{k(k+1)(k+2)}$. Using partial fractions, we can express this as:

$\frac{1}{k(k+1)(k+2)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2}$

To find $A$, $B$, and $C$, multiply both sides by $k(k+1)(k+2)$ and solve for the constants.

$1 = A(k+1)(k+2) + Bk(k+2) + Ck(k+1)$

Expanding this equation:

$1 = A(k^2 + 3k + 2) + B(k^2 + 2k) + C(k^2 + k)$ $1 = (A + B + C)k^2 + (3A + 2B + C)k + (2A)$

Now, by comparing coefficients with the constant 1, we get the following system of equations:

1. $A + B + C = 0$
2. $3A + 2B + C = 0$
3. $2A = 1$

From equation (3), we have $A = \frac{1}{2}$.

Substitute $A = \frac{1}{2}$ into equations (1) and (2):

1. $\frac{1}{2} + B + C = 0 \Rightarrow B + C = -\frac{1}{2}$
2. $3 \times \frac{1}{2} + 2B + C = 0 \Rightarrow \frac{3}{2} + 2B + C = 0 \Rightarrow 2B + C = -\frac{3}{2}$

Now solve this system:

From $B + C = -\frac{1}{2}$, solve for $C = -\frac{1}{2} - B$, and substitute into $2B + C = -\frac{3}{2}$:

$2B + \left( -\frac{1}{2} - B \right) = -\frac{3}{2}$ $2B - \frac{1}{2} - B = -\frac{3}{2}$ $B - \frac{1}{2} = -\frac{3}{2}$ $B = -1$

Now, substitute $B = -1$ into $B + C = -\frac{1}{2}$:

$-1 + C = -\frac{1}{2}$ $C = \frac{1}{2}$

So, we have:

$A = \frac{1}{2}, \quad B = -1, \quad C = \frac{1}{2}$

Thus, the partial fraction decomposition is:

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

Now, let's rewrite the sum $S_n$:

$S_n = \sum_{k=1}^{n} \left( \frac{1/2}{k} - \frac{1}{k+1} + \frac{1/2}{k+2} \right)$

This sum has a telescoping nature, and many terms will cancel out as $n \to \infty$. The remaining terms give us the value of the limit:

$\lim_{n \to \infty} S_n = \boxed{0}$

Would you like more details, or do you have any questions?

Here are 5 related questions:

1. How can partial fraction decomposition help simplify rational expressions?
2. Can you explain the concept of a telescoping series in more detail?
3. What are some other examples of limits involving telescoping series?
4. How would the solution change if the general term were $\frac{1}{k(k+1)}$ instead of $\frac{1}{k(k+1)(k+2)}$?
5. Why does the sum converge to 0 as $n \to \infty$?

Tip: Telescoping series are powerful because they allow you to reduce complicated sums by canceling intermediate terms.