I am trying to solve past year exam problem. I need to show that $\lim _{n\to \infty }\left(\left(1+\frac{1}{n^4}\right)\left(1+\frac{2^4}{n^4}\right)^{1/2}\left(1+\frac{3^4}{n^4}\right)^{1/3}..\left(2\right)^{1/n}\right)= e^{\pi^2/32}$.
I am completely stucked. I need help to solve this problem. Thanks
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Let $${\rm A}=\lim _{n\to \infty }\left(\left(1+\frac{1}{n^4}\right)\left(1+\frac{2^4}{n^4}\right)^{1/2}\left(1+\frac{3^4}{n^4}\right)^{1/3}..\left(2\right)^{1/n}\right)$$
Taking $\ln$ both the sides,
$$\ln {\rm A} =\lim_{n \to \infty} \sum_{r=1}^n \frac 1r \ln \left( 1+ \left( \frac rn \right)^4 \right)$$
$$=\lim_{n \to \infty} \frac 1n \sum_{r=1}^n \frac 1{r/n} \ln \left( 1+ \left( \frac rn \right)^4 \right)= \int_0^1 \frac 1x \ln \left(1+x^4\right) {\rm d} x$$
Continuing from Jaideep Khare's answer, $$\begin{align}\int_0^1 \frac 1x \ln \left(1+x^4\right) d x&=\int_0^1 \frac{1}{x} \left(\sum_{i=1}^\infty(-1)^{i+1}\frac{x^{4i}}{i}\right)\,dx\\ &=\int_0^1 \sum_{i=1}^{\infty} (-1)^{i+1} \frac{x^{4i-1}}{i} \,dx\\ &=\sum_{i=1}^{\infty} (-1)^{i+1} \int_0^1 \frac{x^{4i-1}}{i}\,dx \\ &= \sum_{i=1}^\infty (-1)^{i+1}\frac{1}{4i^2} \\ &=\frac{1}{4}\sum_{i=1}^\infty \frac{(-1)^{i+1}}{i^2} \\ &= \frac{\pi^2}{32}\end{align}$$ The final sum is a standard result (or lemma) related to the Basel problem. Interchange of the sum and limit needs to be justified, I leave to the reader.