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Question: Answered & Verified by Expert
If $\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=k$ then $\log k=$
MathematicsLimitsAP EAMCETAP EAMCET 2019 (21 Apr Shift 1)
Options:
  • A $\log 4+\frac{\pi}{2}-1$
  • B $\log 2+\frac{\pi}{2}+1$
  • C $\log 2+\frac{\pi}{2}-2$
  • D $\log 2+\frac{\pi}{2}-1$
Solution:
2709 Upvotes Verified Answer
The correct answer is: $\log 2+\frac{\pi}{2}-2$
Given,
$$
\begin{aligned}
& k=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n} \\
& \log k=\lim _{n \rightarrow \infty} \log \left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n} \\
= & \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\frac{r^2}{n^2}\right)=\int_0^1 \log \left(1+x^2\right) d x \\
= & {\left[\log \left(1+x^2\right) \cdot x\right]_0^1-\int_0^1 \frac{2 x}{1+x^2} \cdot x d x } \\
= & \log 2-2 \int_0^1 \frac{x^2}{1+x^2} d x=\log 2-2 \int_0^1\left(1-\frac{1}{1+x^2}\right) d x \\
= & \log 2-2\left[x-\tan ^{-1} x\right]_0^1 \\
= & \log 2-2\left[1-\frac{\pi}{4}\right] \\
= & \log 2-2+\frac{\pi}{2} \\
= & \log 2+\frac{\pi}{2}-2
\end{aligned}
$$

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