If ₙ→∞} ≤ft[≤ft(1+1/n²)≤ft(1+2²/n²)·s≤ft(1+n²/n²)]¹/n} = k , then log k =

Correct Answer is : log 2 + π2π/22π - 2

Correct Answer is : log 2 + π2π/22π - 2

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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 76 of 160

Marks: +1, -0

\lim\limits_{n\to\infty}

\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right)\cdots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}

= k , then log k =

Solution:

Consider the expression,

k \;\; =\lim\limits_{x\to\infty}\left[\left(1+\frac{1}{n^{2}}\right)\left(1+\frac{2^{2}}{n^{2}}\right)\cdot s\left(1+\frac{n^{2}}{n^{2}}\right)\right]^{\frac{1}{n}}

\log k \;\; =\lim\limits_{x\to\infty}\log\left[\left(1+\frac{1}{n^{2}}\right)\left(1+\frac{2^{2}}{n^{2}}\right)\cdot s\left(1+\frac{n^{2}}{n^{2}}\right)\right]^{\frac{1}{n}}

\;\; =\lim\limits_{x\to\infty}\frac{1}{n}\sum\limits_{r=1}^{n}\log\left(1+\frac{r^{2}}{n^{2}}\right)

\;\; =\int\limits_{0}^{1}\log(1+x^{2})\,dx

Solve further

\log k \;\; =\left[\log(1+x^{2})\,x\right]_{0}^{1}-\int\limits_{0}^{1}\log(1+x^{2})\,dx

\;\; =\log 2-2\int\limits_{0}^{1}\frac{x^{2}}{1+x^{2}}\,dx

\;\; =\log 2-2\int\limits_{0}^{1}\left(1-\frac{1}{1+x^{2}}\right)dx

\;\; =\log 2-2\left[x-\tan^{-1}x\right]_{0}^{1}

Simplify further

\log k \;\; =\log 2-2\left[1-\frac{\pi}{4}\right]

\;\; =\log 2-2+\frac{\pi}{2}

\;\; =\log 2+\frac{\pi}{2}-2

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