Question

If $U_n = \left(1 + \frac{1}{n^2}\right) \left(1 + \frac{2^2}{n^2}\right)^2 \dots \left(1 + \frac{n^2}{n^2}\right)^n$, then $\lim_{n \to \infty} (U_n)^{-\frac{4}{n^2}}$ is equal to :

• A. $\frac{4}{e}$
• B. $\frac{4}{e^2}$
• C. $\frac{16}{e^2}$
• D. $\frac{e^2}{16}$

Hint:

$\int_1^2 \ln t dt = 2 \ln 2 - 1$ is a very common result in integration-based limit problems. Memorizing it saves time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This limit of a product can be evaluated by converting it to a limit of a sum using logarithms, which then becomes a definite integral.
Step 2: Detailed Explanation:
Let $L = \lim_{n \to \infty} (U_n)^{-4/n^2}$.
$\ln L = \lim_{n \to \infty} -\frac{4}{n^2} \ln U_n = \lim_{n \to \infty} -\frac{4}{n^2} \sum_{r=1}^n r \ln(1 + r^2/n^2)$.
$\ln L = -4 \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n} (\frac{r}{n}) \ln(1 + (\frac{r}{n})^2)$.
This is a Riemann sum for the integral:
$\ln L = -4 \int_0^1 x \ln(1+x^2) dx$.
Let $1+x^2 = t \implies 2x dx = dt$.
$\ln L = -2 \int_1^2 \ln t dt = -2 [t \ln t - t]_1^2$.
$\ln L = -2 [(2 \ln 2 - 2) - (0 - 1)] = -2 [2 \ln 2 - 1] = 2 - 4 \ln 2$.
$\ln L = \ln(e^2) - \ln(2^4) = \ln(e^2/16)$.
$L = e^2/16$.
Step 3: Final Answer:
The limit is $\frac{e^2}{16}$.