If Un=1+1n21+22n22⋯1+n2n2n, then limn→∞ Un−4n2 is equal to

Let y=limn→∞ Un−4n2 y=limn→∞ 1+1n2−4n21+22n2−4n2⋅21+32n2−4n2⋅3 ⋯Taking log on both sides, we get ln y=limn→∞ ∑r=1n −4n2⋅rln⁡1+r2n2 ⇒ln y=limn→∞ ∑-4r=1n 1n⋅rn ln⁡1+r2n2Now, replace limn→∞ Σ→∫ rn→x, 1n→dxLower limit = 0Upper limit = 1∴ ln⁡y=∫01 −4xln⁡1+x2dx Let 1+x2=t⇒ xdx=dt2 When x→0, t→1 and x→1, t→2∴ ln⁡y=∫12 −2ln⁡tdt =−2(tln⁡t−t)12 =−2(2ln⁡2−2+1) =−2(2ln⁡2−1)⇒ ln⁡y=ln⁡116+2⇒ y=116e2

If Un=1+1n21+22n22⋯1+n2n2n, then limn→∞ Un−4n2 is equal to

Let y=limn→∞ Un−4n2 y=limn→∞ 1+1n2−4n21+22n2−4n2⋅21+32n2−4n2⋅3 ⋯Taking log on both sides, we get ln y=limn→∞ ∑r=1n −4n2⋅rln⁡1+r2n2 ⇒ln y=limn→∞ ∑-4r=1n 1n⋅rn ln⁡1+r2n2Now, replace limn→∞ Σ→∫ rn→x, 1n→dxLower limit = 0Upper limit = 1∴ ln⁡y=∫01 −4xln⁡1+x2dx Let 1+x2=t⇒ xdx=dt2 When x→0, t→1 and x→1, t→2∴ ln⁡y=∫12 −2ln⁡tdt =−2(tln⁡t−t)12 =−2(2ln⁡2−2+1) =−2(2ln⁡2−1)⇒ ln⁡y=ln⁡116+2⇒ y=116e2

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

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$16 / e^{2}$

$e^{2} / 16$

$4 / e^{2}$

$4 / e$

answer is A.

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Detailed Solution

$\begin{array}{l} Let y = lim_{n \to \infty} {(U_{n})}^{\frac{- 4}{n^{2}}} \\ y = lim_{n \to \infty} [{(1 + \frac{1}{n^{2}})}^{\frac{- 4}{n^{2}}} {(1 + \frac{2^{2}}{n^{2}})}^{\frac{- 4}{n^{2}} \cdot 2} {(1 + \frac{3^{2}}{n^{2}})}^{\frac{- 4}{n^{2}} \cdot 3} \dots] \end{array}$

Taking log on both sides, we get

$\ln y = lim_{n \to \infty} \sum_{r = 1}^{n} [\frac{- 4}{n^{2}} \cdot rln (1 + \frac{r^{2}}{n^{2}})] \Rightarrow \ln y = lim_{n \to \infty} {\sum (- 4)}_{r = 1}^{n} [\frac{1}{n} \cdot \frac{r}{n} \ln (1 + \frac{r^{2}}{n^{2}})]$

Now, replace $lim_{n \to \infty} Σ \to \int$

$\frac{r}{n} \to x, \frac{1}{n} \to dx$

Lower limit = 0

Upper limit = 1

$\begin{array}{l} ∴ \ln y = \int_{0}^{1} - 4 xln (1 + x^{2}) dx \\ Let 1 + x^{2} = t \\ \Rightarrow xdx = \frac{dt}{2} \\ When x \to 0, t \to 1 \\ and x \to 1, t \to 2 \\ ∴ \ln y = \int_{1}^{2} - 2 \ln tdt \\ = - 2 (tln t - t)_{1}^{2} \\ = - 2 (2 \ln 2 - 2 + 1) \\ = - 2 (2 \ln 2 - 1) \\ \Rightarrow \ln y = \ln \frac{1}{16} + 2 \\ \Rightarrow y = \frac{1}{16} e^{2} \end{array}$