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Recursive Sequence: Prove the Partial Sum < 1

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

Question
A sequence AnA_n is defined by A1=12A_1=\frac12 and An=2n−32nAn−1A_n=\frac{2n-3}{2n}A_{n-1} for n≥2n\ge 2. Then
A∑k=1nAk=1\sum_{k=1}^{n}A_k=1 for all n≥1n\ge 1
B∑k=1nAk>1\sum_{k=1}^{n}A_k>1 for all n≥1n\ge 1
C∑k=1nAk<1\sum_{k=1}^{n}A_k<1 for all n≥1n\ge 1correct
DNone of these
Solution
Step 1: Clear the fraction: 2n An=(2n−3)An−12n\,A_n=(2n-3)A_{n-1}. Write 2n−3=2(n−1)−12n-3=2(n-1)-1:
2n An−2(n−1)An−1=−An−1.2n\,A_n-2(n-1)A_{n-1}=-A_{n-1}.
 Step 2: Sum from k=2k=2 to n+1n+1; the left side telescopes and the right side is −∑k=1nAk-\sum_{k=1}^{n}A_k:
2(n+1)An+1−2A1=−∑k=1nAk ⇒ ∑k=1nAk=1−2(n+1)An+1(A1=12).2(n+1)A_{n+1}-2A_1=-\sum_{k=1}^{n}A_k\ \Rightarrow\ \sum_{k=1}^{n}A_k=1-2(n+1)A_{n+1}\quad(A_1=\tfrac12).
 Step 3: Every An>0⇒2(n+1)An+1>0⇒∑k=1nAk<1A_n>0\Rightarrow 2(n+1)A_{n+1}>0\Rightarrow \sum_{k=1}^{n}A_k<1 for all n≥1.n\ge 1. Correct answer: (3)
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