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

1. Problem Description

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

2. Solution Steps

We can use partial fraction decomposition to rewrite the fraction. We want to find constants $A$ and $B$ such that

$\frac{2}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2}$.

Multiplying by $k(k+2)$ gives

$2 = A(k+2) + Bk$.

Let $k = 0$. Then $2 = A(0+2) + B(0)$, so $2 = 2A$ and $A = 1$.

Let $k = -2$. Then $2 = A(-2+2) + B(-2)$, so $2 = -2B$ and $B = -1$.

Therefore,

$\frac{2}{k(k+2)} = \frac{1}{k} - \frac{1}{k+2}$.

Now we can rewrite the sum as

$\sum_{k=1}^{\infty} \frac{2}{k(k+2)} = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+2}\right)$.

This is a telescoping sum. Let's write out the first few terms:

$(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{4} - \frac{1}{6}) + (\frac{1}{5} - \frac{1}{7}) + \dots$

We can see that the terms $-\frac{1}{3}, -\frac{1}{4}, -\frac{1}{5}, \dots$ will cancel with the terms $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots$

We can write the partial sum $S_n$ as

$S_n = \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+2}\right) = \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{n-1} - \frac{1}{n+1}\right) + \left(\frac{1}{n} - \frac{1}{n+2}\right)$.

The terms that remain are $1, \frac{1}{2}, -\frac{1}{n+1}, -\frac{1}{n+2}$.

So $S_n = 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}$.

As $n \to \infty$, $\frac{1}{n+1} \to 0$ and $\frac{1}{n+2} \to 0$.

Therefore, $\sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+2}\right) = 1 + \frac{1}{2} = \frac{3}{2}$.

3. Final Answer

$\frac{3}{2}$