Limit of a sequence (3)

Let the limit we need to find be $L$ .

$\begin{aligned} L &=\lim_{n \rightarrow \infty} \prod_{k=1}^{n} \sqrt[n]{a(k,n)} \\ &=\lim_{n \rightarrow \infty} \prod_{k=1}^{n} \sqrt[n]{1+\dfrac{k^2}{n^2}} \\ &=\lim_{n \rightarrow \infty} \sqrt[n]{\left(1+\dfrac{1^2}{n^2}\right) \left(1+\dfrac{2^2}{n^2}\right) \ .... \ \left(1+\dfrac{n^2}{n^2}\right)} \end{aligned}$

Taking natural logarithm on both the sides to convert the limit in question into a sum,

$\begin{aligned} \log L &=\lim_{n \rightarrow \infty} \dfrac{1}{n} \log{\left(\left(1+\dfrac{1^2}{n^2}\right) \left(1+\dfrac{2^2}{n^2}\right) \ .... \ \left(1+\dfrac{n^2}{n^2}\right)\right)} \\ &=\lim_{n \rightarrow \infty} \dfrac{1}{n} \displaystyle\sum_{r=1}^{n} \log {\left(1+\dfrac{r^2}{n^2}\right)} \\ &=\lim_{n \rightarrow \infty} \displaystyle\int_{1/n}^{n/n} \log (1+x^2) \ dx \\ &= \displaystyle\int_{0}^{1} \log {(1+x^2)} \ dx \\ \therefore L &=e^{\int_{0}^{1} \log {(1+x^2)} \ dx} \end{aligned}$

Now to find the integral, we use IBP $\rightarrow$

$\begin{aligned} \displaystyle\int_{0}^{1} \log {(1+x^2)} \ dx &= \left. x\log {(1+x^2)} \right|_{0}^{1}-\displaystyle\int_{0}^{1} x \dfrac{2x}{1+x^2} dx \\ &= \log 2-\displaystyle\int_{0}^{1} 2-\dfrac{2}{1+x^2} dx \\ &= \log 2 - \left. 2x-2\tan^{-1} x\right|_{0}^{1}=\log 2+\dfrac{\pi}{2}-2 \end{aligned}$

So, $L=\boxed{2e^{\pi/2-2}} \approx \boxed{1.302}$