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I have been evaluating limits of the Collatz and Waring sequences and have found one strange result (top line). For all of the others, $-\infty, +\infty$ produce equal results. $$ \lim_{n\to -\infty }(3^n-1)^{\frac{1}{ \log (2^n-1)}} =e $$ $$ \lim_{n\to \infty }(3^n-1)^{\frac{1}{ \log (2^n-1)}} =3^{\frac{1}{ \log 2}} $$

What doe this indicate about the sequence? Why $e$? I'm only working with positive numbers, so should I be concerned with $-\infty$?

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  • $\begingroup$ Oh I see the predicament. You are concerned with taking the log of negative numbers and negative numbers raised to non-integer powers? $\endgroup$ – Simply Beautiful Art Oct 15 '16 at 13:24
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Note that we can write

$$\begin{align} \lim_{n\to -\infty}\left((3^n-1)^{1/\log\left(2^n-1\right)}\right)&=\lim_{n\to -\infty}e^{\frac{\log(3^n-1)}{\log(2^n-1)}}\\\\ &=\lim_{n\to -\infty}e^{\frac{\log(1-3^n)+i(2k+1)\pi}{\log(1-2^n)+i(2k+1)\pi}}\\\\ &=e^{\lim_{n\to \infty}\frac{\log(1-3^n)+i(2k+1)\pi}{\log(1-2^n)+i(2k+1)\pi}}\\\\ &=e^{1} \end{align}$$

as was to be shown.

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  • $\begingroup$ $\exp ( \log _{2^n-1}(3^n-1))$ was my construction method, which might have something to do with it. $\endgroup$ – Fred Kline Oct 15 '16 at 14:49
  • $\begingroup$ Yes, that would pose a problem. ;-) $\endgroup$ – Mark Viola Oct 15 '16 at 16:02
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$\log(x)$ may be defined for $x\in\mathbb C$ in the following manner:

$$\log(a+bi)=\log|a+bi|+i\arg(a+bi)$$

where $\arg(a+bi)$ is the argument (angle from the positive real axis) when $a+bi$ is drawn in the complex plane.

We also have Euler's formula for complex exponents, that gives us

$$e^{a+bi}=e^a(\cos(b)+i\sin(b))$$

And,

$$a+bi=|a+bi|e^{i\arg(a+bi)}$$

So that

$$(a+bi)^{c+di}=|a+bi|^{c+di}e^{-d\arg(a+bi)+ci\arg(a+bi)}$$

Depending on what you are doing, extending such operations to complex numbers may be of importance.

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  • $\begingroup$ The argument of a complex number is multi-valued and hence the complex logarithm is likewise. ;-)) $\endgroup$ – Mark Viola Oct 15 '16 at 14:31

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