Question

The value of $\underset{n\to \mathrm{\infty }}{lim}\sum _{r=1}^n{\displaystyle \frac{r^2}{n^3+n^2+r}}$ equals:
A. ${\displaystyle \frac{1}{3}}$
B. ${\displaystyle \frac{1}{2}}$
C. ${\displaystyle \frac{2}{3}}$
D. 1

Answer 1

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

Complete step-by-step answer:
Let’s say that $S_n=\sum _{r=1}^n{\displaystyle \frac{r^2}{n^3+n^2+r}}$ .
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,
⇒ $${\displaystyle \frac{1}{n^3+n^2+n}}\le {\displaystyle \frac{1}{n^3+n^2+r}}\le {\displaystyle \frac{1}{n^3}}$$
Multiply by r2, we get
⇒ $${\displaystyle \frac{r^2}{n^3+n^2+n}}\le {\displaystyle \frac{r^2}{n^3+n^2+r}}\le {\displaystyle \frac{r^2}{n^3}}$$
Taking submission from r=1 to r=n,
⇒ $$\sum _{r=1}^n{\displaystyle \frac{r^2}{n^3+n^2+n}}\le \sum _{r=1}^n{\displaystyle \frac{r^2}{n^3+n^2+r}}\le \sum _{r=1}^n{\displaystyle \frac{r^2}{n^3}}$$
On solving,
⇒ $${\displaystyle \frac{1}{n^3+n^2+n}}\sum _{r=1}^n{r}^2\le S_n\le {\displaystyle \frac{1}{n^3}}\sum _{r=1}^n{r}^2$$
⇒ $${\displaystyle \frac{1}{n^3+n^2+n}}\left[{\displaystyle \frac{n(n+1)(2n+1)}{6}}\right]\le S_n\le {\displaystyle \frac{1}{n^3}}\left[{\displaystyle \frac{n(n+1)(2n+1)}{6}}\right]$$
On applying the limits, we get:
⇒ $${\displaystyle \frac{1}{6}}\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{2n^3+3n^2+n}{n^3+n^2+n}}\right]\le \underset{n\to \mathrm{\infty }}{lim}{S}_n\le {\displaystyle \frac{1}{6}}\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{2n^3+3n^2+n}{n^3}}\right]$$
Dividing the numerator and the denominator by the highest power of n, we get:
⇒ $${\displaystyle \frac{1}{6}}\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{2+\frac{3}{n}+\frac{1}{n^2}}{1+\frac{1}{n}+\frac{1}{n^2}}}\right]\le \underset{n\to \mathrm{\infty }}{lim}{S}_n\le {\displaystyle \frac{1}{6}}\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{2+\frac{3}{n}+\frac{1}{n^2}}{1}}\right]$$
Now, as $n\to \mathrm{\infty },{\displaystyle \frac{1}{n}}\to 0$ .
⇒ $${\displaystyle \frac{1}{6}}\left[{\displaystyle \frac{2+0+0}{1+0+0}}\right]\le \underset{n\to \mathrm{\infty }}{lim}{S}_n\le {\displaystyle \frac{1}{6}}\left[{\displaystyle \frac{2+0+0}{1}}\right]$$
⇒ $${\displaystyle \frac{1}{6}}(2)\le \underset{n\to \mathrm{\infty }}{lim}{S}_n\le {\displaystyle \frac{1}{6}}(2)$$
⇒ $${\displaystyle \frac{1}{3}}\le \underset{n\to \mathrm{\infty }}{lim}{S}_n\le {\displaystyle \frac{1}{3}}$$
Therefore, by using the squeeze theorem, $$\underset{n\to \mathrm{\infty }}{lim}{S}_n=\underset{n\to \mathrm{\infty }}{lim}\sum _{r=1}^n{\displaystyle \frac{r^2}{n^3+n^2+r}}={\displaystyle \frac{1}{3}}$$ .
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.