If U(n) = (1 + \(\frac{1^2}{n^2}\)) \((1 +\frac{2^2}{n^2})^2\)\((1 +\frac{3^2}{n^2})^3\)......\((1 +\frac{n^2}{n^2})^n\), then find \(\lim\limits_{x \to \infty}[U(n)]^{\frac{-4}{n^2}}\)
log (U(n)) = \(\sum_{r=1}^{n}\)log \((1 +\frac{r^2}{n^2})^r\)
= \(\sum_{r=1}^{n}\)log \((1 +\frac{r^2}{n^2})\)
U(n) = \(e^{\sum_{r=1}^{n}}\)r log \((1 +\frac{r^2}{n^2})\)
\(\Big[U(n)\Big]^{\frac{-4}{n^2}}\) = \(e^{\sum_{r=1}^{n}}\)r log \((1 +\frac{r^2}{n^2})\) x \(\frac{-4}{n^2}\)
= \(e^{\sum_{r=1}^{n}}\)r log\((1 +\frac{r^2}{n^2})\) x(-4)\(\frac{r}{n} \times \frac{1}{n}\)
= \(e^{-4 \int_0^1 x log(1+x^2)dx}\)
[Let 1 + x2 = t ⇒ 2xdx = dt]
= \(e^{\frac{-4}{2} \int_1^2 logtdt}\)
\(\Big[U(n)\Big]^{\frac{-4}{n^2}}\) = \(e^{-2(t\;ln\;t-t)_1^2}\)
= e-2[(2 ln 2-2)-(1 ln 1-1)]
= e-2[2 ln 2 - 1]
= e-4 ln 2+2
= \(\frac{e^2}{16}\)