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

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Limits

U_{n} = (1 + \frac{1}{n ^{2}}) (1 + \frac{2 ^{2}}{n ^{2}})^{2} \dots (1 + \frac{n ^{2}}{n ^{2}})^{n}

, then

lim_{n \to \infty} (U_{n})^{\frac{- 4}{n ^{2}}}

is equal to

Answer: (D)

Let

I = lim_{n \to \infty} (U_{n})^{\frac{- 4}{n ^{2}}}

Taking logarithm:

lo g I = n \to \infty lim \frac{- 4}{n ^{2}} r = 1 \sum n r lo g (1 + \frac{r ^{2}}{n ^{2}})

= n \to \infty lim \frac{- 4}{n} r = 1 \sum n \frac{r}{n} lo g (1 + (\frac{r}{n})^{2})

Converting to integral:

lo g I = - 4 \int_{0}^{1} x lo g (1 + x^{2}) d x

Let

t = 1 + x^{2}

, then

d t = 2 x d x

lo g I = - 2 \int_{1}^{2} lo g t d t = - 2 [t lo g t - t]_{1}^{2}

= - 2 [(2 lo g 2 - 2) - (0 - 1)] = - 2 [2 lo g 2 - 1]

= - 2 lo g \frac{4}{e} = lo g \frac{e ^{2}}{16}

Therefore,

I = \frac{e ^{2}}{16}

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option