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Evaluate the limit: $$\lim_{n\rightarrow +\infty}\left [ i^i\left ( 2i \right )^{2i} \cdots\left ( ni \right )^{ni}\right ]$$

where we take into account the branch of the log. function that the argument lies within the interval $(-\pi, \pi]$.

Any suggestions on how to begin?

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  • $\begingroup$ Have you tried rewriting all the powers in terms of the exponential function and the logarithm? $\endgroup$ – Mariano Suárez-Álvarez Mar 7 '15 at 4:17
  • $\begingroup$ That is exactly what I have not done!! And I am feeling embarassed. $\endgroup$ – Tolaso Mar 7 '15 at 4:35
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Since $$(ki)^{ki} = e^{ki\log(ki)} = e^{ki(\log k + i\pi/2)} = e^{-k\pi/2}e^{(k\log k)i}\quad (k = 1,2,\ldots, n)$$ we have

$$\lim_{n\to \infty} i^{i}(2i)^{2i}\ldots (ni)^{ni} = \lim_{n\to \infty} \prod_{k = 1}^n e^{-k\pi/2} e^{(k\log k)i} = \lim_{n\to \infty} e^{-n(n+1)\pi} e^{i\sum_{k = 1}^n k\log k}.$$

Use the fact that $e^{-n(n+1)\pi} \to 0$ as $n \to \infty$ and $|e^{i\sum_{k = 1}^n k\log k}| = 1$ to conclude that $\lim\limits_{n\to \infty} i^i(2i)^{2i}\cdots (ni)^{ni} = 0$.

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  • $\begingroup$ This is slightly more than a suggestion on how to begin... :-) $\endgroup$ – Mariano Suárez-Álvarez Mar 7 '15 at 4:45

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