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 Post subject: Two complex limits
PostPosted: Thu Jul 14, 2016 1:54 pm 
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Choose that branch of the logarithmic function for which the argument function takes its values in \( \displaystyle ( -\pi , \pi ] \).

(1) Compute the limit

\[ \displaystyle \lim_{n \to \infty} \left[ i^{i} (2i)^{2i} \dots (ni)^{ni} \right] \]

(2) Consider the sequence \( (z_{n})_{n \in \mathbb{N}} \) of complex numbers defined by \( z_{n} = \frac{i}{n} \). Show that

\[ \displaystyle \lim_{n \to \infty} \left[ (z_{1})^{z_{1}} \dots (z_{n})^{z_{n}} \right] = 0 \]


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 Post subject: Re: Two complex limits
PostPosted: Thu Jul 14, 2016 1:55 pm 
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Tsakanikas Nickos wrote:
Choose that branch of the logarithmic function for which the argument function takes its values in \( \displaystyle ( -\pi , \pi ] \).

(1) Compute the limit

\[ \displaystyle \lim_{n \to \infty} \left[ i^{i} (2i)^{2i} \dots (ni)^{ni} \right] \]


Since the argument lies within the interval \( (-\pi, \pi] \) we get that:
$$\left ( ki \right )^{ki}=e^{ki\ln (ki)}=e^{ki\left ( \ln k+i\pi/2 \right )}=e^{-k\pi/2}e^{i(k\ln k)}$$

Hence the limit is expressed as:
$$\lim_{n\rightarrow +\infty}\left [ i^i(2i)^{2i}\cdots (ni)^{ni} \right ]=\lim_{n\rightarrow +\infty}\prod_{k=1}^{n}e^{-k\pi/2}e^{ik\ln k}=\lim_{n\rightarrow +\infty}e^{-n(n+1)\pi}e^{i\sum_{k=1}^{n}k \ln k}$$

We note that \( \displaystyle \left | e^{i\sum_{k=1}^{n}k\ln k} \right |\leq1 \) and as \( n \rightarrow +\infty \) we have \( e^{-n(n+1)} \rightarrow 0 \) . Hence the limit is zero.

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