Among the following statements:(S1): lim _{n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For $(S1)$: The sum of the first $n$ even numbers is $2(1+2+3+\ldots+n) = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Thus,$\lim _{n ightarrow \infty} \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For $(S1)$: The sum of the first $n$ even numbers is $2(1+2+3+\ldots+n) = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Thus,$\lim _{n ightarrow \infty} \frac{n(n+1)}{n^2} = \lim _{n ightarrow \infty} \frac{n^2+n}{n^2} = \lim _{n ightarrow \infty} (1 + \frac{1}{n}) = 1$. So,$(S1)$ is true.For $(S2)$: We use the definition of a definite integral as the limit of a sum: $\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(\frac{r}{n}) = \int_{0}^{1} f(x) dx$.Here,$\lim _{n ightarrow \infty} \frac{1}{n^{16}} \sum_{r=1}^{n} r^{15} = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} (\frac{r}{n})^{15} = \int_{0}^{1} x^{15} dx$.Evaluating the integral: $\int_{0}^{1} x^{15} dx = [\frac{x^{16}}{16}]_{0}^{1} = \frac{1}{16}$. So,$(S2)$ is true.

Among the following statements:
$(S1): \lim _{n \rightarrow \infty} \frac{1}{n^2}(2+4+6+\ldots+2n)=1$
$(S2): \lim _{n \rightarrow \infty} \frac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+\ldots+n^{15})=\frac{1}{16}$

Similar Questions

If $\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=k$,then $\log k=$

If $a = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{2n}{n^2+k^2}$ and $f(x) = \sqrt{\frac{1-\cos x}{1+\cos x}}$,$x \in (0, 1)$,then:

$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\frac{3}{n^2} \sec ^2 \frac{9}{n^2}+\ldots+\frac{n}{n^2} \sec ^2 \frac{n^2}{n^2}\right]=$

$\lim _{n \rightarrow \infty}\left(\frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\ldots+\frac{n}{n^2+n^2}\right)=$

Evaluate $\int_{-1}^{2}(7 x-5) d x$ as a limit of sums.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Among the following statements:$(S1): \lim _{n \rightarrow \infty} \frac{1}{n^2}(2+4+6+\ldots+2n)=1$$(S2): \lim _{n \rightarrow \infty} \frac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+\ldots+n^{15})=\frac{1}{16}$