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2024 PEA Q2

Suppose SnS_n is defined as follows for every positive integer n≥2n\ge 2:

Sn=(1−122) ⁣(1−132)⋯ ⁣(1−1n2). S_n=\left(1-\tfrac{1}{2^2}\right)\!\left(1-\tfrac{1}{3^2}\right)\cdots\!\left(1-\tfrac{1}{n^2}\right).

The value of lim⁡n→∞Sn\displaystyle\lim_{n\to\infty}S_n is

Reveal answer and solution

Answer

B

Solution

  1. 1

    Each factor factors as

  2. 2
    1−1k2=(k−1)(k+1)k2. 1-\frac{1}{k^2}=\frac{(k-1)(k+1)}{k^2}.
  3. 3

    Therefore

  4. 4
    Sn=∏k=2n(k−1)(k+1)k2=(∏k=2nk−1k) ⁣(∏k=2nk+1k)=1n⋅n+12=n+12n. S_n=\prod_{k=2}^{n}\frac{(k-1)(k+1)}{k^2} =\left(\prod_{k=2}^{n}\frac{k-1}{k}\right)\!\left(\prod_{k=2}^{n}\frac{k+1}{k}\right) =\frac{1}{n}\cdot\frac{n+1}{2} =\frac{n+1}{2n}.
  5. 5

    Hence lim⁡n→∞Sn=12\lim_{n\to\infty}S_n=\tfrac{1}{2}.

Answer structure / marking notes

No major trap beyond standard calculation care.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2024/ISI_MSQE_PEA_2024_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.