Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
If $\lim_{n \to \infty} \left( \frac{(n^3+1)(n^3+2^3)(n^3+3^3)\ldots(n^3+n^3)}{n^{3n}} \right)^{1/n} = 4e^{3/4}e^{-b}$ (where $a, b \in \mathbb{N}$) then $a + b$ is ______.
Step-by-Step Solution
Key Concept: Converting a discrete sum into a Riemann integral limit and evaluating it using calculus techniques.
Given $\ln P = \frac{1}{n}[\ln(n^3+1) + \ln(n^3+2^3) + ... + \ln(n^3+n^3)] - 3n\ln n$, we simplify to find $T_r = \frac{\ln[1+(r/n)^3]}{n}$. Taking the limit as a Riemann sum: $S = \lim_{n \to \infty} \sum_{r=1}^{n} \ln[1+(r/n)^3] = \int_0^1 \ln(1+x^3)dx + \int_0^1 \ln(x^2-x+1)dx$. After evaluating these integrals using integration by parts and substitution, we obtain $\ln P = \ln 4 + \frac{\pi}{\sqrt{3}} - 3$, giving $P = 4e^{\pi/\sqrt{3}-3}$.
Correct Answer: I need to find the values of $a$ and $b$ from the given limit expression.
Given that:
$$\lim_{n \to \infty} \left( \frac{(n^3+1)(n^3+2^3)(n^3+3^3)\ldots(n^3+n^3)}{n