Among : (S1) : \(\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{n}^{2}}(2+4+6+\ldots \ldots \ldots+2 \mathrm{n})=1\) (S2) : \(\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{n}^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots \ldots+\mathrm{n}^{15}\right)=\frac{1}{16}\)

(A) Both (S1) and (S2) are true

(B) Both (S1) and (S2) are false

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Limits

Among :
(S1) :

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

(S2) :

lim_{n \to \infty} \frac{1}{n ^{16}} (1^{15} + 2^{15} + 3^{15} + \dots\dots\dots + n^{15}) = \frac{1}{16}

Answer: (A)

For (S1):

\sum_{k = 1}^{n} 2 k = 2 \cdot \frac{n ( n + 1 )}{2} = n (n + 1)

. So

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

. True. For (S2):

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{1}{16}

. True. Both statements are true.

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Single correct option