Lim [1/(1×2×3) + ...+ 1/(n×(n+1)+(n+2))]

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.

Explore the limit of a series involving partial fractions and telescoping series. This problem includes concepts like series summation, partial fraction decomposition, and how to simplify complex expressions. Suitable for high school students in Grades 11-12.

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