Integral Calculus> What is ((Lim n tends to infinity)(summat...

What is ((Lim n tends to infinity)(summation from r=1 to n)((r)/(n^2 +n + r)))

Aniket Mohanty , 8 Years ago

Grade 12th pass

2 Answers

Nishant Vora

Last Activity: 8 Years ago

Please check the question it should be

(r)/(n^2 +n * r)))

If this is the question then we can solve by applying definite integration as a limit of sum

Thanks

jagdish singh singh

Last Activity: 8 Years ago

$\hspace{-0.7 cm}$ Using $n^2<n^2+n+r<(n+1)^2\;\forall r=1,2,3,4,........,n$\\\\ $\frac{r}{(n+1)^2}<\frac{r}{n^2+n+r}\leq \frac{r}{n^2},$ When $n\rightarrow \infty$\\\\ So $\lim_{r\rightarrow \infty} \sum^{n}_{r=1}\frac{r}{(n+1)^2}<\lim_{r\rightarrow \infty} \sum^{n}_{r=1}\frac{r}{n^2+n+r}\leq \lim_{r\rightarrow \infty} \sum^{n}_{r=1}\frac{r}{n^2},$\\\\ So $\lim_{r\rightarrow \infty}\frac{n(n+1)}{2(n+1)^2}\leq \lim_{r\rightarrow \infty}\sum^{n}_{r=1}\frac{r}{n^2+n+r}\leq \lim_{r\rightarrow \infty}\frac{n(n+1)}{2n^2}$\\\\ So Using Sandwitch Theorem, We get $\lim_{r\rightarrow \infty} \sum^{n}_{r=1}\frac{r}{n^2+n+r} = \frac{1}{2}$$