Among (S1) : lim n 1 n 2 ( 2 + 4 + 6 + … + 2 n ) = 1 (S2) : lim n 1 n 16 ( 1 15 + 2 15 + 3 15 + … + n 15 ) = 1 16 [2023]

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Solution

Q.
Among

$(S1) : \lim_{n \to \infty} \frac{1}{n^{2}} (2 + 4 + 6 + \dots + 2 n) = 1$

$(S2) : \lim_{n \to \infty} \frac{1}{n^{16}} (1^{15} + 2^{15} + 3^{15} + \dots + n^{15}) = \frac{1}{16}$ [2023]

1 Only (S1) is true

2 Both (S1) and (S2) are false

3 Only (S2) is true

4 Both (S1) and (S2) are true

Ans.
(4)

$(S1):$ $\lim_{n \to \infty} \frac{1}{n^{2}} (2 + 4 + 6 + \dots + 2 n)$

$= \lim_{n \to \infty} \frac{1}{n^{2}} \times 2 (1 + 2 + 3 + \dots + n)$

$= \lim_{n \to \infty} \frac{1}{n^{2}} \times 2 \times \frac{n (n + 1)}{2} = \lim_{n \to \infty} (1 + \frac{1}{n}) = 1$

$∴$ $(S 1) is true.$

$(S2):$ $\lim_{n \to \infty} \frac{1}{n^{16}} (1^{15} + 2^{15} + 3^{15} + \dots + n^{15}) = \lim_{n \to \infty} \frac{1}{n^{16}} \sum_{r = 1}^{n} r^{15}$

$= \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{15} = \int_{0}^{1} x^{15} d x = | \frac{x^{16}}{16} |_{0}^{1} = \frac{1}{16}$

$∴$ $(S 2) is also true.$

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