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MCQMediumJEE Main 2023 · 13 April, Shift 1Limits

Mathematics Question from JEE Main 2023 · 13 April, Shift 1

Among

(S1) lim⁡n→∞1n2(2+4+6+⋯+2n)=1\lim_{n \to \infty} \frac{1}{n^{2}} \left( 2 + 4 + 6 + \dots + 2n \right) = 1

(S2) lim⁡n→∞1n16(115+215+315+⋯+n15)=116\lim_{n \to \infty} \frac{1}{n^{16}} \left( 1^{15} + 2^{15} + 3^{15} + \dots + n^{15} \right) = \frac{1}{16}

  • A

    Only (S1) is true

  • B

    Both (S1) and (S2) are true

  • C

    Both (S1) and (S2) are false

  • D

    Only (S2) is true

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