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How to calculate : $$\lim_{n\to\infty}\frac{1}{n^4}\prod_{i=1}^{2n}\left(n^2+i^2\right)^{\frac{1}{n}}$$

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Take logarithm of both sides maybe. This transforms product into sum and powers into products. – mbaitoff Feb 1 '13 at 12:54
Hint: Rewrite the logarithm of this as a Riemann sum... – achille hui Feb 1 '13 at 12:55
up vote 4 down vote accepted

Pull the $n$ out of the product to get

$$L = \lim_{n \rightarrow \infty} \prod_{i=1}^{2 n} \left ( 1+\frac{i^2}{n^2} \right )^\frac{1}{n} $$

Rewrite as

$$\log{L} = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{2 n} \left ( 1+\frac{i^2}{n^2} \right )$$

This is a Riemann sum whose limit is

$$\begin{align}\log{L} &= \int_0^2 dx \: \log{(1+x^2)}\\ &= [x \log{(1+x^2)}]_0^2 - 2 \int_0^2 dx \: \frac{x^2}{1+x^2} \\ &= 2 \log{5} + 2 \arctan{2} - 4 \end{align}$$

Then the limit $L$ is

$$L = 25 e^{-2 (2-\arctan{2})} $$

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In the second line, I think the right hand side should be $\displaystyle\lim_{n\to\infty}\frac{1}{n}\dots$ as opposed to $\displaystyle\frac{1}{n}\lim_{n\to\infty}\dots$ – Michael Albanese Feb 1 '13 at 13:53
Yes, thanks for catching that goof. – Ron Gordon Feb 1 '13 at 14:12

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