Step 1: Recall the alkane test formula.
Any saturated alkane must fit the pattern $\text{C}_n\text{H}_{2n+2}$, so its hydrogen count should always be two more than twice the carbon count.
Step 2: Apply this test to each formula.
For $\text{CH}_4$, $n = 1$ gives $2(1)+2 = 4$, which matches. For $\text{C}_2\text{H}_6$, $n=2$ gives $2(2)+2=6$, which matches. For $\text{C}_3\text{H}_8$, $n=3$ gives $2(3)+2=8$, which matches.
Step 3: Now test $\text{C}_4\text{H}_8$. \[ 2(4) + 2 = 10 \neq 8 \] The hydrogen count actually fits $\text{C}_n\text{H}_{2n}$ instead, which is the alkene pattern, telling us this molecule has a double bond hiding in it (it is butene).
Step 4: Draw the conclusion.
Since the first three obey the alkane rule and the fourth does not, $\text{C}_4\text{H}_8$ is the odd one out and belongs to a different homologous series. \[ \boxed{\text{C}_4\text{H}_8} \]