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To solve the given series and prove that its sum is \( S = \frac{1}{4} \), let’s carefully analyze and compute step by step.

The series is:
\[
\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}
\]

### Step 1: Decompose the fraction
The term \(\frac{1}{n(n+1)(n+2)}\) can be decomposed into partial fractions:
\[
\frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}
\]
Multiply through by \(n(n+1)(n+2)\) to find \(A\), \(B\), and \(C\):
\[
1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1)
\]
Expanding:
\[
1 = A(n^2 + 3n + 2) + B(n^2 + 2n) + C(n^2 + n)
\]
\[
1 = (A + B + C)n^2 + (3A + 2B + C)n + (2A)
\]

Equating coefficients:
1. \(A + B + C = 0\)
2. \(3A + 2B + C = 0\)
3. \(2A = 1 \implies A = \frac{1}{2}\)

From \(A = \frac{1}{2}\), substitute into the first equation:
\[
\frac{1}{2} + B + C = 0 \implies B + C = -\frac{1}{2}
\]
Substitute \(A = \frac{1}{2}\) into the second equation:
\[
3\left(\frac{1}{2}\right) + 2B + C = 0 \implies \frac{3}{2} + 2B + C = 0
\]
Simplify:
\[
2B + C = -\frac{3}{2}
\]

Now solve \(B + C = -\frac{1}{2}\) and \(2B + C = -\frac{3}{2}\):
\[
(2B + C) - (B + C) = -\frac{3}{2} - \left(-\frac{1}{2}\right)
\]
\[
B = -1
\]
Substitute \(B = -1\) into \(B + C = -\frac{1}{2}\):
\[
-1 + C = -\frac{1}{2} \implies C = \frac{1}{2}
\]

Thus:
\[
A = \frac{1}{2}, \, B = -1, \, C = \frac{1}{2}
\]

### Step 2: Rewrite the series
Using the partial fraction decomposition:
\[
\frac{1}{n(n+1)(n+2)} = \frac{\frac{1}{2}}{n} - \frac{1}{n+1} + \frac{\frac{1}{2}}{n+2}
\]

The series becomes:
\[
\sum_{n=1}^\infty \left(\frac{\frac{1}{2}}{n} - \frac{1}{n+1} + \frac{\frac{1}{2}}{n+2}\right)
\]
\[
= \frac{1}{2}\sum_{n=1}^\infty \frac{1}{n} - \sum_{n=1}^\infty \frac{1}{n+1} + \frac{1}{2}\sum_{n=1}^\infty \frac{1}{n+2}
\]

### Step 3: Simplify the telescoping series
Shift the indices for \(\sum \frac{1}{n+1}\) and \(\sum \frac{1}{n+2}\):
1. For \(\sum_{n=1}^\infty \frac{1}{n+1}\), let \(k = n+1\):
   \[
   \sum_{n=1}^\infty \frac{1}{n+1} = \sum_{k=2}^\infty \frac{1}{k}
   \]
2. For \(\sum_{n=1}^\infty \frac{1}{n+2}\), let \(m = n+2\):
   \[
   \sum_{n=1}^\infty \frac{1}{n+2} = \sum_{m=3}^\infty \frac{1}{m}
   \]

Rewrite the series:
\[
\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n} - \sum_{k=2}^\infty \frac{1}{k} + \frac{1}{2} \sum_{m=3}^\infty \frac{1}{m}
\]

Observe how the terms cancel telescopically:
\[
\frac{1}{2} \left(1 + \frac{1}{2} + \frac{1}{3} + \dots\right) - \left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots\right) + \frac{1}{2} \left(\frac{1}{3} + \frac{1}{4} + \dots\right)
\]

Combine similar terms:
\[
\frac{1}{2}(1) + \left(-\frac{1}{2} + \frac{1}{2}\right)\left(\frac{1}{2} + \frac{1}{3} + \dots\right)
\]

The result simplifies to:
\[
S = \frac{1}{4}
\]

### Final Answer:
The sum of the series is:
\[
S = \frac{1}{4}
\]

Would you like a deeper explanation of any step or additional examples? 

#### 5 Related Questions:
1. How do you solve general partial fraction decomposition problems?
2. What are the conditions for a series to telescope?
3. Can this method be applied to a different series, such as \( \sum \frac{1}{n(n+1)} \)?
4. How do we rigorously prove convergence for this type of series?
5. What role does shifting indices play in telescoping series?

**Tip**: For telescoping series, always look for cancellations by rewriting terms systematically!

Arătați că suma seriei \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} \) este \( S = \frac{1}{4} \).

Explore the step-by-step solution for proving that the sum of the infinite series \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} \) equals \( \frac{1}{4} \). Learn techniques such as partial fraction decomposition and telescoping series. Suitable for advanced mathematics students.

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