1. Assume there exists an infinite geometric subsequence in the arithmetic sequence.

Let the arithmetic sequence be $a, a+b, a+2b, \ldots$.

Suppose there exists an infinite geometric subsequence. Let three consecutive terms of this geometric subsequence be $a + xb, a + yb, a + zb$ where $x, y, z$ are integers with $x < y < z$.

2. Establish the common ratio of the geometric subsequence.

The common ratio $k$ of the geometric subsequence is given by:
$$k = \frac{a + yb}{a + xb} = \frac{a + zb}{a + yb}$$

3. **Solve for $k$ in terms of $x, y, z$.**

From the first equality:
$$k = \frac{a + yb}{a + xb}$$
Rearrange to get:
$$k(a + xb) = a + yb$$
$$ka + kxb = a + yb$$
$$ka - a = yb - kxb$$
$$a(k - 1) = b(y - kx)$$
$$\frac{a}{b} = \frac{y - kx}{k - 1}$$

4. **Conclude that $\frac{a}{b}$ is rational.**

Since $x, y, z$ are integers, $k$ must be rational. Therefore, $\frac{y - kx}{k - 1}$ is rational, implying $\frac{a}{b}$ is rational.

5. **Assume $\frac{a}{b}$ is rational and show the existence of an infinite geometric subsequence.**

Let $\frac{a}{b} = \frac{p}{q}$ where $p$ and $q$ are integers. Then the arithmetic sequence can be written as:
$$a = \frac{bp}{q}, \quad a + b = \frac{bp}{q} + b, \quad a + 2b = \frac{bp}{q} + 2b, \ldots$$

6. Construct the geometric subsequence.

Consider the terms:
$$\frac{bp}{q}, \quad \frac{bp}{q} + b, \quad \frac{bp}{q} + 2b, \ldots$$
Factor out $\frac{b}{q}$:
$$\frac{b}{q} \left( p, p + q, p + 2q, \ldots \right)$$
This sequence contains all terms of the form $\frac{b}{q} \cdot p(q+1)^n$ for non-negative integers $n$, which forms a geometric sequence with common ratio $q+1$.

Conclusion:
$$\boxed{\text{The sequence contains an infinite geometric subsequence if and only if } \frac{a}{b} \text{ is rational.}}$$