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Step-by-Step Solution
Step 1: Understand the Given Conditions
We are given two conditions on the function $f$:
$f(x + y) = f(x)\,f(y)$ for all natural numbers $x, y$.
$\displaystyle \sum_{x=1}^{\infty} f(x) = 2$.
Step 2: Derive Expressions for Small Arguments
Using the functional equation $f(x + y) = f(x)\,f(y)$, we can find $f(2)$, $f(3)$, and $f(4)$ in terms of $f(1)$.
Set $x = 1, y = 1$:
\[
f(2) = f(1+1) = f(1)\,f(1) = \bigl(f(1)\bigr)^2.
\]
Set $x = 2, y = 1$:
\[
f(3) = f(2+1) = f(2)\,f(1) = \bigl(f(1)\bigr)^2 \cdot f(1) = \bigl(f(1)\bigr)^3.
\]
Set $x = 2, y = 2$:
\[
f(4) = f(2+2) = f(2)\,f(2) = \bigl(f(1)\bigr)^2 \cdot \bigl(f(1)\bigr)^2 = \bigl(f(1)\bigr)^4.
\]
Step 3: Use the Infinite Series Condition
We know
\[
\sum_{x=1}^{\infty} f(x) = f(1) + f(2) + f(3) + \dots = 2.
\]
Substituting the expressions for $f(2)$, $f(3)$, etc., in terms of $f(1)$, we get:
\[
f(1) + \bigl(f(1)\bigr)^2 + \bigl(f(1)\bigr)^3 + \dots = 2.
\]
This is an infinite geometric series with first term $f(1)$ and common ratio $f(1)$.
Step 4: Solve for $f(1)$
The sum of an infinite geometric series with first term $a$ and common ratio $r$ (where $|r| < 1$) is given by
\[
\frac{a}{1 - r}.
\]
Here, $a = f(1)$ and $r = f(1)$. Thus,
\[
2 = \frac{f(1)}{1 - f(1)}.
\]
Solving for $f(1)$:
\[
2 - 2\,f(1) = f(1) \quad \Longrightarrow \quad 2 = 3\,f(1) \quad \Longrightarrow \quad f(1) = \frac{2}{3}.
\]
Step 5: Find $f(2)$ and $f(4)$
\[
f(2) = \bigl(f(1)\bigr)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}.
\]
\[
f(4) = \bigl(f(1)\bigr)^4 = \left(\frac{2}{3}\right)^4 = \frac{16}{81}.
\]
Step 6: Compute the Requested Ratio
We need to calculate
\[
\frac{f(4)}{f(2)} = \frac{\frac{16}{81}}{\frac{4}{9}} = \frac{16}{81} \times \frac{9}{4} = \frac{16 \times 9}{81 \times 4} = \frac{144}{324} = \frac{4}{9}.
\]
Hence,
\[
\frac{f(4)}{f(2)} = \frac{4}{9}.
\]
Final Answer
The value of $f(4)/f(2)$ is $\boxed{\frac{4}{9}}$.
