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Sequences & Series: Distinct Odd Natural Numbers Divisible Prime Greater Less

JEE Maths question with a full step-by-step solution.

Question
If a1,a2,,ana_1, a_2, \ldots, a_n are distinct odd natural numbers not divisible by any prime greater than 55, then
1a1+1a2++1an\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}
is less than
A158\frac{15}{8}correct
B168\frac{16}{8}
C815\frac{8}{15}
D154\frac{15}{4}
Solution
Step 1: Each aia_i is odd and uses only 33 and 55, so ai=3r5sa_i=3^r 5^s with r,s0.r,s\ge 0. Step 2: The reciprocals are therefore bounded by the product of two geometric series:
1a1++1an<(1+13+132+)(1+15+152+)=11131115=3254=158.\frac{1}{a_1}+\cdots+\frac{1}{a_n}<\left(1+\frac13+\frac{1}{3^2}+\cdots\right)\left(1+\frac15+\frac{1}{5^2}+\cdots\right)=\frac{1}{1-\frac13}\cdot\frac{1}{1-\frac15}=\frac32\cdot\frac54=\frac{15}{8}.
Step 3: Only finitely many terms are taken, so the sum is strictly less than 158.\dfrac{15}{8}. Correct answer: (1)
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