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

The value of $\mathop {\text{Lim}}\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^{4n} {\frac{{\sqrt n }}{{\sqrt r {{\left( {\,3\sqrt r + 4\sqrt n \,} \right)}^2}}}} $ is equal to

$\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n}{(n+1)^{2}}+\frac{n}{(n+2)^{2}}+\ldots+\frac{n}{(2 n-1)^{2}}\right]$ is equal to ...... .

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

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2-1^2}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)$ is equal to

The value of $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^3}{r^4+n^4}$ is

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}$