$\lim _{n \rightarrow \infty} \left[ \frac{n}{n^2+1^2} + \frac{n}{n^2+2^2} + \dots + \frac{n}{n^2+n^2} \right]$ ની કિંમત શોધો.

  • A
    $\frac{\pi}{4}$
  • B
    $\log 2$
  • C
    શૂન્ય
  • D
    $1$

Explore More

Similar Questions

$a \in R, |a| > 1$ માટે,ધારો કે $\lim _{n \rightarrow \infty} \left( \frac{1+\sqrt[3]{2}+\ldots+\sqrt[3]{n}}{n^{7/3} \left( \frac{1}{(an+1)^2} + \frac{1}{(an+2)^2} + \ldots + \frac{1}{(an+n)^2} \right)} \right) = 54$. તો $a$ ની શક્ય કિંમત(ઓ) શોધો:
$(1) 8$ $(2) -9$ $(3) -6$ $(4) 7$

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 0}^n {\frac{n}{{{{\left( {2r + n} \right)}^2}}}} $ ની કિંમત શોધો.

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

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \dots + \frac{1}{{2n}}} \right] = $

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo