We are asked to evaluate the infinite sum $\sum_{k=2}^{\infty} (\frac{1}{k} - \frac{1}{k-1})$.

1. Problem Description

We are asked to evaluate the infinite sum $\sum_{k=2}^{\infty} (\frac{1}{k} - \frac{1}{k-1})$.

2. Solution Steps

Let $S_n$ be the partial sum:

$S_n = \sum_{k=2}^{n} (\frac{1}{k} - \frac{1}{k-1})$

Expanding the sum gives:

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

Notice that this is a telescoping sum, where most terms cancel out. We have:

$S_n = -\frac{1}{1} + (\frac{1}{2} - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{3}) + ... + (\frac{1}{n-1} - \frac{1}{n-1}) + \frac{1}{n}$

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

Now, we take the limit as $n$ approaches infinity to find the value of the infinite sum:

$\sum_{k=2}^{\infty} (\frac{1}{k} - \frac{1}{k-1}) = \lim_{n \to \infty} S_n = \lim_{n \to \infty} (-1 + \frac{1}{n})$

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

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

3. Final Answer

The final answer is -1.