The problem shows a geometric sequence $a_1, a_2, a_3, \ldots, a_n$ where:

• $a_2 = \frac{4}{25}$
• $a_4 = \frac{16}{625}$
• $r > 0$ is the common ratio
• You are asked to find the infinite sum $\sum_{n=1}^{\infty} a_n$

Step 1: Find the common ratio $r$

Since this is a geometric sequence, the relationship between terms is given by:

$a_n = a_1 \cdot r^{n-1}$

Using the known values of $a_2$ and $a_4$:

$a_2 = a_1 \cdot r = \frac{4}{25}$

$a_4 = a_1 \cdot r^3 = \frac{16}{625}$

We now have the system of equations:

$a_1 \cdot r = \frac{4}{25}$ $a_1 \cdot r^3 = \frac{16}{625}$

Dividing the second equation by the first to eliminate $a_1$:

$\frac{a_1 \cdot r^3}{a_1 \cdot r} = \frac{\frac{16}{625}}{\frac{4}{25}}$

Simplifying:

$r^2 = \frac{16}{625} \times \frac{25}{4} = \frac{16 \times 25}{625 \times 4} = \frac{400}{2500} = \frac{4}{25}$

Thus:

$r = \frac{2}{5}$

Step 2: Find $a_1$

Now substitute $r = \frac{2}{5}$ into the equation for $a_2$:

$a_1 \cdot \frac{2}{5} = \frac{4}{25}$

Solving for $a_1$:

$a_1 = \frac{4}{25} \cdot \frac{5}{2} = \frac{20}{50} = \frac{2}{5}$

Step 3: Find the infinite sum

The sum of an infinite geometric series is given by:

$S = \frac{a_1}{1 - r}$

Substitute $a_1 = \frac{2}{5}$ and $r = \frac{2}{5}$:

$S = \frac{\frac{2}{5}}{1 - \frac{2}{5}} = \frac{\frac{2}{5}}{\frac{3}{5}} = \frac{2}{5} \cdot \frac{5}{3} = \frac{2}{3}$

Final Answer:

The infinite sum of the geometric sequence is:

$\boxed{\frac{2}{3}}$

Would you like further explanation on any step? Here are some related questions:

1. How do you derive the sum formula for an infinite geometric series?
2. Can the sum formula be applied if $r \geq 1$?
3. What happens to the series if $r < 0$?
4. How would the sum change if the first term were different?
5. Could you solve for any general term $a_n$ in this sequence?

Tip: Always verify that the common ratio $r$ satisfies $|r| < 1$ when calculating the infinite sum, otherwise, the series diverges.