Question

S₁ = $\sum_{r=1}^{n}r$, S₂ = $\sum_{r=1}^{n}r^2$ and S₃ = $\sum_{r=1}^{n}r^3$, then the value of $\lim_{n \to \infty} \frac{s_1(1+\frac{s_3}{4})}{s_2^2}$ is

• A. $\frac{9}{16}$
• B. $\frac{9}{2}$
• C. $\frac{9}{32}$
• D. $\frac{9}{8}$

Answer 1

Answer: (C)

$\frac{9}{32}$

Answer 2

For large $n$, we use the asymptotic formulas:

$$\begin{aligned} S_1 &= \sum_{r=1}^{n} r \approx \frac{n^2}{2}, \\ S_2 &= \sum_{r=1}^{n} r^2 \approx \frac{n^3}{3}, \\ S_3 &= \sum_{r=1}^{n} r^3 \approx \left(\frac{n^2}{2}\right)^2 = \frac{n^4}{4}. \end{aligned}$$

The given expression is:

$$\lim_{n \to \infty} \frac{S_1\left(1+\frac{S_3}{4}\right)}{S_2^2}.$$

Substitute the asymptotic values:

$$1+\frac{S_3}{4} \approx 1+\frac{n^4}{16} \sim \frac{n^4}{16} \quad (\text{since } \frac{n^4}{16} \gg 1).$$

Now, the numerator becomes:

$$S_1\left(1+\frac{S_3}{4}\right) \sim \frac{n^2}{2} \cdot \frac{n^4}{16} = \frac{n^6}{32}.$$

And the denominator:

$$S_2^2 \sim \left(\frac{n^3}{3}\right)^2 = \frac{n^6}{9}.$$

Thus, the limit is:

$$\lim_{n \to \infty} \frac{\frac{n^6}{32}}{\frac{n^6}{9}} = \frac{9}{32}.$$