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The given sum is:

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

We can break this down using partial fraction decomposition. Let's decompose the fraction:

\[
\frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}
\]

Multiplying both sides by \(n(n+1)(n+2)\), we get:

\[
1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1)
\]

Now, expand the terms:

\[
1 = A(n^2 + 3n + 2) + B(n^2 + 2n) + C(n^2 + n)
\]

\[
1 = (A + B + C)n^2 + (3A + 2B + C)n + (2A)
\]

Equating the coefficients on both sides, we get the following system of equations:

1. \(A + B + C = 0\)
2. \(3A + 2B + C = 0\)
3. \(2A = 1\)

From the third equation, \(A = \frac{1}{2}\).

Substitute \(A = \frac{1}{2}\) into the first and second equations:

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

Now, solve these two equations:

- \(B + C = -\frac{1}{2}\)
- \(2B + C = -\frac{3}{2}\)

Subtract the first equation from the second:

\[
(2B + C) - (B + C) = -\frac{3}{2} - \left(-\frac{1}{2}\right)
\]

\[
B = -1
\]

Now substitute \(B = -1\) into \(B + C = -\frac{1}{2}\):

\[
-1 + C = -\frac{1}{2} \quad \Rightarrow \quad C = \frac{1}{2}
\]

Thus, the partial fraction decomposition is:

\[
\frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2}
\]

Now we can rewrite the sum as:

\[
\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} = \sum_{n=1}^{\infty} \left( \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} \right)
\]

This is a telescoping series, where many terms will cancel out. Expanding the first few terms, we see the pattern of cancellation. After the cancellation, the remaining terms leave a finite result. Specifically, this sum converges to:

\[
\boxed{\frac{1}{4}}
\]

Would you like further details on any part of the process?

---

Here are some related questions to expand your understanding:

1. How does partial fraction decomposition work in general?
2. Why do we use telescoping series to simplify sums?
3. How can we generalize this method for more complex rational functions?
4. What conditions must be met for an infinite series to converge?
5. What are other common techniques for summing infinite series?

**Tip**: When dealing with infinite series, always check if you can apply known formulas or simplifications such as telescoping or geometric series.

Explore how to find the sum of an infinite series \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} \) using partial fraction decomposition and telescoping techniques. The step-by-step solution simplifies the series and demonstrates convergence to a constant.

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