Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • <img alt="\mathrm{\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r}{\left(n^2+n+r\right)} \text { equals : }}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bn%20%5C

\mathrm{\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r}{\left(n^2+n+r\right)} \text { equals : }}

Option: 1

0


Option: 2

\frac{1}{3}


Option: 3

\frac{1}{2}


Option: 4

1


Answers (1)

best_answer


\mathrm{\begin{aligned} f(n) & =\frac{1}{\left(n^2+n+1\right)}+\frac{2}{\left(n^2+n+2\right)}+\ldots+\frac{n}{\left(n^2+n+n\right)} \\ g(n) & =\frac{1}{\left(n^2+n+n\right)}+\frac{2}{\left(n^2+n+n\right)}+\ldots+\frac{n}{\left(n^2+n+n\right)} \\ & =\frac{n(n+1)}{2\left(n^2+2 n\right)} \end{aligned}}\mathrm{g(n)<f(n) Similarly, \begin{aligned} h(n) & =\frac{1}{\left(n^2+n+1\right)}+\frac{2}{\left(n^2+n+1\right)}+\ldots .+\frac{n}{\left(n^2+n+1\right)} \\ & =\frac{n(n+1)}{2\left(n^2+n+1\right)} \end{aligned}}
\mathrm{ \begin{aligned} g(n) & <f(n)<h(n) \\ \quad \lim _{n \rightarrow \infty} g(n) & <\lim _{n \rightarrow \infty} f(n)<\lim _{n \rightarrow \infty} h(n) \\ \Rightarrow \quad \frac{1}{2} & <\lim _{n \rightarrow \infty} f(n)<\frac{1}{2} \end{aligned} Using Sandwich theorem. \lim _{n \rightarrow \infty} f(n)=\frac{1}{2} }

Posted by

Anam Khan

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Maharashtra to introduce policy for regulating coaching centres: Report
anam-khan

JEE Main 2026: BTech civil engineering cut-offs at NITs show steady demand after years of decline
anam-khan

JEE Main 2026 Registration LIVE: NTA session 1 application link active at jeemain.nta.nic.in; apply by Nov 27

Latest Question