Question

The value of $\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{\dfrac{{{r}^{2}}}{{{n}^{3}}+{{n}^{2}}+r}}$ equals:
A. $\dfrac{1}{3}$
B. $\dfrac{1}{2}$
C. $\dfrac{2}{3}$
D. 1

Answer 1

The squeeze theorem states that if we define functions such that h(x) ≤ f(x) ≤ g(x) and if $\underset{x\to a}{\mathop{\lim }}\,h(x)=\underset{x\to a}{\mathop{\lim }}\,g(x)=L$ , then $\underset{x\to a}{\mathop{\lim }}\,f(x)=L$ .
The sum $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\dfrac{n(n+1)(2n+1)}{6}$ .

Complete step-by-step answer:
Let’s say that ${{S}_{n}}=\sum\limits_{r=1}^{n}{\dfrac{{{r}^{2}}}{{{n}^{3}}+{{n}^{2}}+r}}$ .
Since 0 < r ≤ n, we can say that:
${{n}^{3}}+{{n}^{2}}+n\ge {{n}^{3}}+{{n}^{2}}+r\ge {{n}^{3}}$
After reciprocal them, sign of inequality changes,
⇒ $\dfrac{1}{{{n}^{3}}+{{n}^{2}}+n}\le \dfrac{1}{{{n}^{3}}+{{n}^{2}}+r}\le \dfrac{1}{{{n}^{3}}}$
Multiply by r2, we get
⇒ $\dfrac{{{r}^{2}}}{{{n}^{3}}+{{n}^{2}}+n}\le \dfrac{{{r}^{2}}}{{{n}^{3}}+{{n}^{2}}+r}\le \dfrac{{{r}^{2}}}{{{n}^{3}}}$
Taking submission from r=1 to r=n,
⇒ $\sum\limits_{r=1}^{n}{\dfrac{{{r}^{2}}}{{{n}^{3}}+{{n}^{2}}+n}}\le \sum\limits_{r=1}^{n}{\dfrac{{{r}^{2}}}{{{n}^{3}}+{{n}^{2}}+r}}\le \sum\limits_{r=1}^{n}{\dfrac{{{r}^{2}}}{{{n}^{3}}}}$
On solving,
⇒ $\dfrac{1}{{{n}^{3}}+{{n}^{2}}+n}\sum\limits_{r=1}^{n}{{{r}^{2}}}\le {{S}_{n}}\le \dfrac{1}{{{n}^{3}}}\sum\limits_{r=1}^{n}{{{r}^{2}}}$
⇒ $\dfrac{1}{{{n}^{3}}+{{n}^{2}}+n}\left[ \dfrac{n(n+1)(2n+1)}{6} \right]\le {{S}_{n}}\le \dfrac{1}{{{n}^{3}}}\left[ \dfrac{n(n+1)(2n+1)}{6} \right]$
On applying the limits, we get:
⇒ $\dfrac{1}{6}\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{2{{n}^{3}}+3{{n}^{2}}+n}{{{n}^{3}}+{{n}^{2}}+n} \right]\le \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}\le \dfrac{1}{6}\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{2{{n}^{3}}+3{{n}^{2}}+n}{{{n}^{3}}} \right]$
Dividing the numerator and the denominator by the highest power of n, we get:
⇒ $\dfrac{1}{6}\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{2+\tfrac{3}{n}+\tfrac{1}{{{n}^{2}}}}{1+\tfrac{1}{n}+\tfrac{1}{{{n}^{2}}}} \right]\le \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}\le \dfrac{1}{6}\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{2+\tfrac{3}{n}+\tfrac{1}{{{n}^{2}}}}{1} \right]$
Now, as $n\to \infty ,\dfrac{1}{n}\to 0$ .
⇒ $\dfrac{1}{6}\left[ \dfrac{2+0+0}{1+0+0} \right]\le \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}\le \dfrac{1}{6}\left[ \dfrac{2+0+0}{1} \right]$
⇒ $\dfrac{1}{6}(2)\le \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}\le \dfrac{1}{6}(2)$
⇒ $\dfrac{1}{3}\le \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}\le \dfrac{1}{3}$
Therefore, by using the squeeze theorem, $\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{\dfrac{{{r}^{2}}}{{{n}^{3}}+{{n}^{2}}+r}}=\dfrac{1}{3}$ .
The correct answer option is A.

Note: The squeeze theorem is typically used to confirm the limit of a function via comparison with two other functions whose limits are known or easily computed.