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 Post subject: Two complex limitsPosted: Thu Jul 14, 2016 1:54 pm
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Joined: Tue Nov 10, 2015 8:25 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 limitsPosted: Thu Jul 14, 2016 1:55 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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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