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Definite Integrals Part 2

Section: Mathematics

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Question : 61 of 100

Marks: +1, -0

If

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

\lim\limits_{n\rightarrow\infty}(U_{n})^{-\frac{4}{n^{2}}}

[27 Aug 2021 Shift 1]

$\frac{e^{2}}{16}$
$\frac{4}{e}$
$\frac{16}{e^{2}}$
$\frac{4}{e^{2}}$

Solution:

Let

y=\lim\limits_{n\rightarrow\infty}(U_{n})^{-\frac{4}{n^{2}}}

y=\lim\limits_{n\rightarrow\infty}\left[\left(1+\frac{1}{n^{2}}\right)^{-\frac{4}{n^{2}}}\left(1+\frac{2^{2}}{n^{2}}\right)^{-\frac{4}{n^{2}}\cdot 2}\left(1+\frac{3^{2}}{n^{2}}\right)^{-\frac{4}{n^{2}}\cdot 3}\dots\right]

Taking log on both sides, we get

\ln y=\lim\limits_{n\rightarrow\infty}\sum\limits_{r=1}^{n}\left[-\frac{4}{n^{2}}\cdot r\ln\left(1+\frac{r^{2}}{n^{2}}\right)\right]

Now, replace

\lim\limits_{n\rightarrow\infty}\Sigma\rightarrow\int

\frac{r}{n}\rightarrow x,\ \frac{1}{n}\rightarrow dx

Lower limit = 0

Upper limit = 1

\therefore \ln y = \int\limits_{0}^{1} -4x \ln(1+x^{2})\,dx

Let

1+x^{2}=t

\Rightarrow x\,dx = \frac{dt}{2}

When x→ 0, t → 1

and x → 1, t → 2

\therefore \ln y = \int\limits_{1}^{2} -2 \ln t\,dt

=-2\left(t\ln t - t\right)\Big|_{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}

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