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!