We are tasked with proving that the infinite sum $\sum_{x=1}^{\infty} \frac{1}{x(x+1)} = 1$.

Step-by-Step Solution

Step 1: Decompose the general term

We start by simplifying the general term $\frac{1}{x(x+1)}$. This can be decomposed using partial fraction decomposition:

$\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$

Multiplying both sides by $x(x+1)$, we get:

$1 = A(x+1) + Bx$

Simplifying:

$1 = A(x+1) + Bx = Ax + A + Bx = (A+B)x + A$

For this equation to hold for all $x$, we must have:

• $A + B = 0$
• $A = 1$

From $A = 1$, it follows that $B = -1$.

Thus, the decomposition is:

$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$

Step 2: Write the sum using the decomposition

Now we rewrite the infinite sum:

$\sum_{x=1}^{\infty} \frac{1}{x(x+1)} = \sum_{x=1}^{\infty} \left( \frac{1}{x} - \frac{1}{x+1} \right)$

This is a telescoping series, where most terms cancel out.

Step 3: Simplify the telescoping series

Writing out the first few terms:

$\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots$

Notice that all terms cancel out except for the first term $\frac{1}{1}$, leaving:

$1$

Step 4: Conclusion

Thus, the sum of the series is:

$\sum_{x=1}^{\infty} \frac{1}{x(x+1)} = 1$

This completes the proof.

---

Would you like further clarification or additional details? Here are 5 related questions:

1. How do telescoping series work in more complicated sums?
2. Can partial fraction decomposition be used for other types of series?
3. What are other examples of series that converge to simple values?
4. How can this technique be applied to sums that don't start at 1?
5. What other methods exist for proving the convergence of an infinite series?

Tip: Always look for patterns like telescoping behavior when summing series—it often simplifies the problem greatly!