Answer
\[
\boxed{g_k = \frac{1}{2^k - 1}.}
\]
Work Step by Step
First, let’s compute the first few terms by hand to spot a pattern:
\(g_1 = 1.\)
\(g_2 = \frac{g_1}{g_1 + 2} = \frac{1}{1+2} = \frac{1}{3}.\)
\(g_3 = \frac{g_2}{g_2 + 2} = \frac{\tfrac{1}{3}}{\tfrac{1}{3}+2} = \frac{\tfrac{1}{3}}{\tfrac{7}{3}} = \frac{1}{7}.\)
\(g_4 = \frac{g_3}{g_3 + 2} = \frac{\tfrac{1}{7}}{\tfrac{1}{7}+2} = \frac{\tfrac{1}{7}}{\tfrac{15}{7}} = \frac{1}{15}.\)
We see a clear pattern:
\[
g_1 = \frac{1}{1}, \quad g_2 = \frac{1}{3}, \quad g_3 = \frac{1}{7}, \quad g_4 = \frac{1}{15}, \quad \dots
\]
Notice that
\[
1,\, 3,\, 7,\, 15, \dots
\]
are all of the form \(2^k - 1\). Hence we guess
\[
g_k = \frac{1}{2^k - 1}.
\]
Let’s verify it satisfies the recurrence:
\[
\text{Given } g_{k-1} = \frac{1}{2^{k-1} - 1},\quad
\text{then }
\frac{g_{k-1}}{g_{k-1} + 2}
= \frac{\frac{1}{2^{k-1}-1}}{\frac{1}{2^{k-1}-1} + 2}
= \frac{\tfrac{1}{2^{k-1}-1}}{\tfrac{1 + 2(2^{k-1}-1)}{2^{k-1}-1}}
= \frac{1}{1 + 2(2^{k-1}-1)}
= \frac{1}{2^k - 1}.
\]
This is exactly the form we guessed for \(g_k\). It also matches the initial condition \(g_1 = 1/(2^1 - 1) = 1\).
