If we want. We can try to treat this as a uniqueness claim for extending hyperoperators to complex numbers. That if we find a formula for \( \vartheta \) that satisfies recursion for natural numbers it should also satisfy recursion for complex numbers. Ergo; uniqueness for how hyper operators behave for complex numbers. It's not very lucid yet. Again; one step at a time.
EDIT:
evaluation of limit
\( \begin{eqnarray*}
\vartheta_n(s) &=& \frac{s}{n} e^{e^s-e^n} \prod_{k=1}^{\infty}\frac{1-\frac{s}{k}}{1-\frac{n}{k}}e^{ \frac{n - s}{k}}\\ &=& \frac{s e^{e^s} \prod_{k=1}^{\infty}(1-\frac{s}{k})e^{\frac{-s}{k}}}{n e^{e^n} \prod_{k=1}^{\infty}(1-\frac{n}{k})e^{\frac{-n}{k}}} = \frac{\psi(s)}{\psi(n)}\\
\end{eqnarray*} \)
\( \begin{eqnarray*}
f(t) &=& \ln(x\,\, \bigtriangleup_{t}\,\, y)\\
&=& \ln(x+y) \frac{\psi(t)}{t} + \sum_{k=1}^{\infty} \frac{\psi(t)}{\psi(k)} \ln(x\,\, \bigtriangleup_{k}\,\, y)\\
&=& \psi(t) \cdot (\frac{\ln(x+y)}{t} + \sum_{k=1}^{\infty} \frac{\ln(x\,\, \bigtriangleup_{k}\,\,y)}{\psi(k)} )
\end{eqnarray*} \)
Therefore we can rephrase the limit as; keeping in mind the first term \( \ln(x+y)/t \) disappears:
\( \begin{eqnarray*}
L &=& \lim_{t \to \infty} | \frac{f(t+1)}{f(t)} \cdot \frac{\vartheta_{t+1}(s)}{\vartheta_{t}(s)}|\\
&=& \lim_{t \to \infty} | \frac{\psi(t+1) \vartheta_{t+1}(s) } {\psi(t) \vartheta_{t}(s)} \cdot \frac{ \frac{\ln(x+y)}{t+1} + \sum_{k=1}^{\infty} \frac{f(k)}{\psi(k)}}{\frac{\ln(x+y)}{t} + \sum_{k=1}^{\infty} \frac{f(k)}{\psi(k)} } |\\
&=&1
\end{eqnarray*} \)
Which means convergence is inconclusive. Using this method generally we will have this result; so long as \( \vartheta \) remains a quotient. Anyone know any other convergence tests? I'm pretty sure this outlaws the root test as inconclusive as well. The integral test? I'll keep looking... Maybe I made a mistake. I tried using the integral test and I got divergence. The methods were a little questionable however. I'll hold out on posting that just yet.
EDIT:
evaluation of limit
\( \begin{eqnarray*}
\vartheta_n(s) &=& \frac{s}{n} e^{e^s-e^n} \prod_{k=1}^{\infty}\frac{1-\frac{s}{k}}{1-\frac{n}{k}}e^{ \frac{n - s}{k}}\\ &=& \frac{s e^{e^s} \prod_{k=1}^{\infty}(1-\frac{s}{k})e^{\frac{-s}{k}}}{n e^{e^n} \prod_{k=1}^{\infty}(1-\frac{n}{k})e^{\frac{-n}{k}}} = \frac{\psi(s)}{\psi(n)}\\
\end{eqnarray*} \)
\( \begin{eqnarray*}
f(t) &=& \ln(x\,\, \bigtriangleup_{t}\,\, y)\\
&=& \ln(x+y) \frac{\psi(t)}{t} + \sum_{k=1}^{\infty} \frac{\psi(t)}{\psi(k)} \ln(x\,\, \bigtriangleup_{k}\,\, y)\\
&=& \psi(t) \cdot (\frac{\ln(x+y)}{t} + \sum_{k=1}^{\infty} \frac{\ln(x\,\, \bigtriangleup_{k}\,\,y)}{\psi(k)} )
\end{eqnarray*} \)
Therefore we can rephrase the limit as; keeping in mind the first term \( \ln(x+y)/t \) disappears:
\( \begin{eqnarray*}
L &=& \lim_{t \to \infty} | \frac{f(t+1)}{f(t)} \cdot \frac{\vartheta_{t+1}(s)}{\vartheta_{t}(s)}|\\
&=& \lim_{t \to \infty} | \frac{\psi(t+1) \vartheta_{t+1}(s) } {\psi(t) \vartheta_{t}(s)} \cdot \frac{ \frac{\ln(x+y)}{t+1} + \sum_{k=1}^{\infty} \frac{f(k)}{\psi(k)}}{\frac{\ln(x+y)}{t} + \sum_{k=1}^{\infty} \frac{f(k)}{\psi(k)} } |\\
&=&1
\end{eqnarray*} \)
Which means convergence is inconclusive. Using this method generally we will have this result; so long as \( \vartheta \) remains a quotient. Anyone know any other convergence tests? I'm pretty sure this outlaws the root test as inconclusive as well. The integral test? I'll keep looking... Maybe I made a mistake. I tried using the integral test and I got divergence. The methods were a little questionable however. I'll hold out on posting that just yet.
