The problem is to evaluate the infinite sum $\sum_{k=1}^{\infty} (\frac{1}{k} - \frac{1}{k+1})$.

1. Problem Description

The problem is to evaluate the infinite sum $\sum_{k=1}^{\infty} (\frac{1}{k} - \frac{1}{k+1})$.

2. Solution Steps

This is a telescoping series. Let $S_n$ be the $n$-th partial sum of the series.

$$S_n = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right)$$

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

The intermediate terms cancel out, so we are left with:

$$S_n = 1 - \frac{1}{n+1}$$

The infinite sum is the limit of the partial sums as $n$ approaches infinity:

$$\sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right)$$

Since $\lim_{n \to \infty} \frac{1}{n+1} = 0$, we have:

$$\lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1 - 0 = 1$$

3. Final Answer

1