We need to find the value of the infinite sum $\sum_{k=2}^{\infty} (\frac{1}{k} - \frac{1}{k-1})$.

1. Problem Description

We need to find the value of the infinite sum $\sum_{k=2}^{\infty} (\frac{1}{k} - \frac{1}{k-1})$.

2. Solution Steps

The given sum is a telescoping sum. Let's write out the first few terms to see the pattern.

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

We observe that most of the terms cancel out. Specifically, the $\frac{1}{2}$ term cancels with the $-\frac{1}{2}$ term, the $\frac{1}{3}$ term cancels with the $-\frac{1}{3}$ term, and so on. The only terms that remain are the first term $-\frac{1}{1}$ and the last term $\frac{1}{n}$.

Thus, the partial sum $S_n$ is given by:

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

Now, to find the value of the infinite sum, we need to find the limit of the partial sum $S_n$ as $n$ approaches infinity.

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

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

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

3. Final Answer

$-1$