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Recursive Sequence: Prove the Partial Sum < 1

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

Question

A sequence

A_n

is defined by

A_1=\frac12

and

A_n=\frac{2n-3}{2n}A_{n-1}

for

n\ge 2

. Then

\sum_{k=1}^{n}A_k=1

for all

n\ge 1

\sum_{k=1}^{n}A_k>1

for all

n\ge 1

\sum_{k=1}^{n}A_k<1

for all

n\ge 1

correct

DNone of these

Solution

Step 1: Clear the fraction:

2n\,A_n=(2n-3)A_{n-1}

. Write

2n-3=2(n-1)-1

2n\,A_n-2(n-1)A_{n-1}=-A_{n-1}.

Step 2: Sum from

k=2

n+1

; the left side telescopes and the right side is

-\sum_{k=1}^{n}A_k

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

A_n>0\Rightarrow 2(n+1)A_{n+1}>0\Rightarrow \sum_{k=1}^{n}A_k<1

for all

n\ge 1.

Correct answer: (3)

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