If U_n=_r=1^n(1+(r^2)/(n^2))^r, then _n(U_n)^-4/n^2 is equal to

Question

If $U_n=\displaystyle\prod_{r=1}^n\left(1+\frac{r^2}{n^2}ight)^{r}$, then $\displaystyle\lim_{n	o\infty}(U_n)^{-4/n^2}$ is equal to

Accepted Answer

Let $y=\displaystyle\lim_{n\to\infty}(U_n)^{-4/n^2}$. Taking log:
$$\ln y=\lim_{n\to\infty}\sum_{r=1}^n\left(-\frac{4}{n^2}\right)r\ln\left(1+\frac{r^2}{n^2}\right)$$

Converting to an integral with $r/n\to x$:
$$\ln y=\int_0^1(-4x)\ln(1+x^2)\,dx$$

Let $1+x^2=t\Rightarrow x\,dx=\dfrac{dt}{2}$; limits $t:1\to2$:
$$\ln y=\int_1^2(-2\ln t)\,dt=-2[t\ln t-t]_1^2=-2(2\ln2-2+1)=-2(2\ln2-1)=2-4\ln2$$

$$\Rightarrow \ln y=\ln\frac1{16}+2\ \Rightarrow\ y=\frac1{16}e^2=\frac{e^2}{16}$$

Solutions · 1