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Question: The value of \[\mathop {\lim }\limits_{n \to \infty } \sum\nolimits_{r = 1}^n {\dfrac{{{r^2}}}{{{n^3...

The value of lim⁡n→∞∑r=1nr2n3+n2+r\mathop {\lim }\limits_{n \to \infty } \sum\nolimits_{r = 1}^n {\dfrac{{{r^2}}}{{{n^3} + {n^2} + r}}} equals
A.13\dfrac{1}{3}
B.12\dfrac{1}{2}
C.23\dfrac{2}{3}
D.1

Explanation

Solution

Here we will use Squeeze theorem in which we will find the upper and lower bounds of the given function. And then we will determine the value of the given function.

Complete step-by-step answer:
First it is given that,
f(n)=lim⁡n→∞∑r=1nr2n3+n2+rf\left( n \right) = \mathop {\lim }\limits_{n \to \infty } \sum\nolimits_{r = 1}^n {\dfrac{{{r^2}}}{{{n^3} + {n^2} + r}}}
Now let g(n)g\left( n \right) and h(n)h\left( n \right) be the lower and upper bounds of the given function. If we observe the limits of the given function then lower bound is 1 and upper bound is n.
Now consider
g(n)=1n3+n2+n+4n3+n2+n+.......+n2n3+n2+ng\left( n \right) = \dfrac{1}{{{n^3} + {n^2} + n}} + \dfrac{4}{{{n^3} + {n^2} + n}} + ....... + \dfrac{{{n^2}}}{{{n^3} + {n^2} + n}}
⇒g(n)=1+4+....n2n3+n2+n\Rightarrow g\left( n \right) = \dfrac{{1 + 4 + ....{n^2}}}{{{n^3} + {n^2} + n}}
Sum of squares of first n natural numbers is given by ⇒n(n+1)(2n+1)6 \Rightarrow \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}
⇒g(n)=n(n+1)(2n+1)6(n3+n2+n)\Rightarrow g\left( n \right) = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6\left( {{n^3} + {n^2} + n} \right)}}
Taking n common from denominator and canceling it with n from numerator,
⇒g(n)=(n+1)(2n+1)6(n2+n+1)\Rightarrow g\left( n \right) = \dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6\left( {{n^2} + n + 1} \right)}}
Now we will find the value of h(n)h\left( n \right).
h(n)=1n3+n2+1+4n3+n2+1+.......+n2n3+n2+1h\left( n \right) = \dfrac{1}{{{n^3} + {n^2} + 1}} + \dfrac{4}{{{n^3} + {n^2} + 1}} + ....... + \dfrac{{{n^2}}}{{{n^3} + {n^2} + 1}}
⇒h(n)=1+4+....n2n3+n2+1\Rightarrow h\left( n \right) = \dfrac{{1 + 4 + ....{n^2}}}{{{n^3} + {n^2} + 1}}
Again sum of squares of first n natural numbers is given by ⇒n(n+1)(2n+1)6 \Rightarrow \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}
⇒h(n)=n(n+1)(2n+1)6(n3+n2+1)\Rightarrow h\left( n \right) = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6\left( {{n^3} + {n^2} + 1} \right)}}
Here we found the value of h(n)h\left( n \right).
If we compared we will get that g(n)<f(n)<h(n)g\left( n \right) < f\left( n \right) < h\left( n \right)
So,
lim⁡n→∞g(n)=lim⁡n→∞(n+1)(2n+1)6(n2+n+1)\mathop {\lim }\limits_{n \to \infty } g\left( n \right) = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6\left( {{n^2} + n + 1} \right)}}
Multiplying terms of numerator
⇒lim⁡n→∞2n2+n+2n+16(n2+n+1)\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{2{n^2} + n + 2n + 1}}{{6\left( {{n^2} + n + 1} \right)}}
⇒16lim⁡n→∞2n2+3n+1(n2+n+1)\Rightarrow \dfrac{1}{6}\mathop {\lim }\limits_{n \to \infty } \dfrac{{2{n^2} + 3n + 1}}{{\left( {{n^2} + n + 1} \right)}}
Dividing the terms by n2{n^2}
⇒16lim⁡n→∞2+3n+1n21+1n+1n2\Rightarrow \dfrac{1}{6}\mathop {\lim }\limits_{n \to \infty } \dfrac{{2 + \dfrac{3}{n} + \dfrac{1}{{{n^2}}}}}{{1 + \dfrac{1}{n} + \dfrac{1}{{{n^2}}}}}
Applying the limit value
⇒16lim⁡n→∞2+0+01+0+0\Rightarrow \dfrac{1}{6}\mathop {\lim }\limits_{n \to \infty } \dfrac{{2 + 0 + 0}}{{1 + 0 + 0}}

⇒16×2 ⇒13  \Rightarrow \dfrac{1}{6} \times 2 \\\ \Rightarrow \dfrac{1}{3} \\\

So by Squeeze theorem we can find f(n)f\left( n \right)
lim⁡n→∞f(n)=13\mathop {\lim }\limits_{n \to \infty } f(n) = \dfrac{1}{3}
Hence option A is the correct answer.

Note: Squeeze theorem is used to confirm the limit of a function by comparing that function with two other functions with upper and lower bound of the limit.