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  • If  <img alt="\mathrm{\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r}{n^2+n+r}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7

If  \mathrm{\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r}{n^2+n+r}}  equals to \mathrm{\frac{3}{a}} , then value of  \mathrm{{a}}  is ________.

Option: 1

6


Option: 2

2


Option: 3

0


Option: 4

4


Answers (1)

best_answer

Let \mathrm{f(n)=\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\ldots+\frac{n}{n^2+n+n}}

Consider, \mathrm{g(n)=\frac{1}{n^2+n+n}+\frac{2}{n^2+n+n}+\ldots+\frac{n}{n^2+n+n}}

\mathrm{ =\frac{1+2+3+\ldots+n}{n^2+2 n}=\frac{n(n+1)}{2\left(n^2+2 n\right)} }

\mathrm{ g(n)<f(n) }                              ..........(i)

Similarly,

\mathrm{ h(n)=\frac{1}{n^2+n+1}+\frac{2}{n^2+n+1}+\ldots+\frac{n}{n^2+n+1} \\ }

\mathrm{ \therefore \quad f(n)<h(n) }                 ...........(ii)

But \mathrm{ \lim _{n \rightarrow \infty} g(n)=\lim _{n \rightarrow \infty} h(n)=\frac{1}{2} }

Hence, using Sandwich theorem

\mathrm{ \lim _{n \rightarrow \infty} f(n)=1 / 2 }


 

Posted by

HARSH KANKARIA

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