The value of lim _{n rightarrow infty} prod_{ | vedclass

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Question

(D) Let $y = \lim _{n \rightarrow \infty} \prod_{r=1}^n \left(1+\frac{r^2}{n^2}\right)^{\frac{2 r}{n^2}}$.Taking the natural logarithm on both sides:$

Vedclass · Accepted Answer

$\frac{4}{e}$

Vedclass · Answer

(D) Let $y = \lim _{n \rightarrow \infty} \prod_{r=1}^n \left(1+\frac{r^2}{n^2}\right)^{\frac{2 r}{n^2}}$.Taking the natural logarithm on both sides:$\ln y = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{2 r}{n^2} \ln \left(1+\frac{r^2}{n^2}\right)$.This is a Riemann sum, which can be written as a definite integral:$\ln y = \int_0^1 2x \ln(1+x^2) dx$.Let $t = 1+x^2$, then $dt = 2x dx$. When $x=0, t=1$ and when $x=1, t=2$.$\ln y = \int_1^2 \ln(t) dt$.Using the integration formula $\int \ln(t) dt = t \ln(t) - t$:$\ln y = [t \ln(t) - t]_1^2 = (2 \ln 2 - 2) - (1 \ln 1 - 1) = 2 \ln 2 - 2 + 1 = \ln(4) - 1 = \ln(4) - \ln(e) = \ln \left(\frac{4}{e}\right)$.Therefore, $y = \frac{4}{e}$.

The value of $\lim _{n \rightarrow \infty} \prod_{r=1}^n\left(1+\frac{r^2}{n^2}\right)^{\frac{2 r}{n^2}}$ is equal to

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