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1984

JEE Main 1984 Mathematics question on Limits Continuity and Differentiability

lim⁡n→∞(11−n2+21−n2+…+n1−n2)\lim_{n \to \infty} \left( \frac{1}{1-n^2} + \frac{2}{1-n^2} + \ldots + \frac{n}{1-n^2} \right) is equal to
  1. A.
    0
  2. B.
    −12-\frac{1}{2}
  3. C.
    12\frac{1}{2}
  4. D.
    None of these
Show answer and solution

Correct option: B

Solution

The sum equals [1+2+...+n]/(1-n²) = [n(n+1)/2] / (1-n²) = n(n+1) / [2(1-n²)].

Since 1-n² = -(n²-1) = -(n-1)(n+1), this simplifies to n(n+1) / [-2(n-1)(n+1)] = -n / [2(n-1)].

As n→∞, -n/[2(n-1)] → -1/2.

[1984]

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