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The MRB constant is defined at http://mathworld.wolfram.com/MRBConstant.html. After a lot of looking I found a connection between the MRB constant and applied math:

The MRB constant is $$\sum^\infty_{k=1} (-1)^k\left(k^{1/k}-1\right),$$ and that $k^{1/k}-1$ is the interest rate to multiply an investment $k$ times in $k$ periods -- as well as other growth models involving the more general expression $(1+k)^n$ -- since $\left(\left(k^{1/k}-1\right)+1\right)^k|_{k\in \mathbb{Z^+}}=k.\text{ and }\left(\left(k^{1/n}-1\right)+1\right)^n|_{n\in \mathbb{Z^+}}=k.$

Couldn't we say the result of summing with alternating signs the interest rate to multiply an investment $k$ times in $k$ periods (or the equivalent growth model) could be the end "growth" rate resulting from growth, following decay, following growth, ad infinitum?

Given that the formula for the MRB constant does have this one application, what are some of those applications it can do some work in ( if not really be useful in since here, could we have $A_t = P*(t^{1/t}-1)=P*(e^{\frac {\log(t)} {t}}-1)$ )?

The idea of summing may present a small hurdle, but if we can first establish the applications for $k^{1/k}\text{ or }(k^{1/k}-1)\text{ then }(-1)^k (k^{1/k}-1)$, (noting that he MRB constant is also known as the upper limit point of the sequence of partial sums defined by $S_{n}=\sum^{n}_{k=1} (-1)^k k^{1/k}$), we could finally see which ones of the models best have an application with the summing as I tried to apply in the above paragraph that starts with "Couldn't we say..."

I try to always give credit to anyone that helps me with my research -- and thanks!

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    $\begingroup$ Is there a reason to expect some application? Also, are any other mathematical properties known? $\endgroup$ – Dark Malthorp Jul 17 '18 at 16:18
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    $\begingroup$ I presented my best argument for a growth-rate application in my question, As for other properties, there is the geometry of the MRB constant -- see vixra.org/abs/1609.0082 . $\endgroup$ – Marvin Ray Burns Jul 17 '18 at 16:26
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    $\begingroup$ Tbh, I think you might be too excited about having a constant named after you. Most obscure mathematical constants don't really have an application other than their specific definition, and I don't see any reason why this one should be any different. That being said, I am trying to come up with something for you because I also love obscure constants $\endgroup$ – Dark Malthorp Jul 18 '18 at 18:41
  • $\begingroup$ I know that I'm too exited, (since 1994 when I started) but the last time I did something about it I almost died. So I guess I'm stuck in MANIC CITY! Since I lost religion, I think of MRB as my key to living on, forever! (at least something to be remembered by). $\endgroup$ – Marvin Ray Burns Jul 18 '18 at 21:12
  • $\begingroup$ @DarkMalthorp ,have you seen ebyte.it/library/educards/constants/MathConstants.pdf . $\endgroup$ – Marvin Ray Burns Jul 18 '18 at 21:25
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The Abel-Plana formula, which I am quite fond of, gives an integral for the MRB constant:\begin{eqnarray} C_{MRB} &=& \sum_{n=0}^\infty (-1)^n\left(1 - (n+1)^{\frac{1}{n+1}}\right) \\&=& \frac{i}{2} \int_0^\infty \frac{(-i t+1)^{\frac{1}{-i t+1}} - (i t+1)^{\frac{1}{i t+1}}}{\sinh{\pi t}} dt \\&=& \int_0^\infty\frac{\Im\left((1+i t)^{\frac{1}{1+i t}}\right)}{\sinh \pi t} dt\\&=&\int_0^\infty \frac{(\sqrt{x^2+1})^{1/(x^2+1)} \exp\left(\frac{x\arctan{x}}{x^2+1}\right)\sin\left(\frac{\arctan(x) - x\log\sqrt{x^2+1}}{x^2+1}\right)}{\sinh \pi x}dx \end{eqnarray} which is a big mess, but this integral converges much faster than the series form since the integrand decays exponentially, so it might be numerically easier to compute. Substituting $x=\tan \theta$, it simplifies to the proper integral $$ C_{MRB} = \int_0^{\frac{\pi}2} \frac{\exp\left(\theta\sin\theta\cos\theta\right)\sin\left(\theta\cos^2\theta + \cos\theta\sin\theta\log\cos\theta\right)}{\left(\cos\theta\right)^{2+\cos^2\theta}\sinh(\pi\tan\theta)}d\theta $$

You could also get a similar but uglier formula using Abel-Plana on $C_{MRB} = \sum_{k=1}^{\infty} \left[(2k)^{1/(2k)} - (2k-1)^{1/(2k-1)}\right]$ instead.

I was briefly looking at the zeta-like function $\mathcal{Z} (z)=\sum_{k=1}^\infty (-1)^k \left(k^{z/k} - 1\right)$, but I couldn't find any relation of $\mathcal{Z}$ to any named functions. Since the series converges for all $z$, $\mathcal{Z}$ is an entire function. Other than that, here's the Taylor series and a generalization of the integral above:\begin{eqnarray} \mathcal{Z}(z) &=& \sum_{k=0}^\infty \frac{(-1)^{k+1}\eta^{(k)}(k)}{k!} z^k \\&=& \int_0^\infty\frac{(\sqrt{x^2+1})^{z/(x^2+1)} \exp\left(\frac{x\arctan{x}}{x^2+1}z\right)\sin\left(\frac{\arctan(x) - x\log\sqrt{x^2+1}}{x^2+1}z\right)}{\sinh \pi x}dx \end{eqnarray} where $\eta$ is the Dirichlet eta function. Both should be valid for all $z\in\mathbb{C}$.

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    $\begingroup$ Dark, I'll look carefully at these. Thank you soooooooooooo much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! $\endgroup$ – Marvin Ray Burns Jan 13 at 20:35
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    $\begingroup$ I referenced your answer on the MRB constant record page on WC. Previously, I tried to incorporate that integral transformation, but couldn't get the same value for CMRB. Now that I understand it, I do get the constant from the integral, as you can see in the WC post. I must have been applying the transformation wrongly. Thanks again! $\endgroup$ – Marvin Ray Burns Jan 15 at 9:55
  • $\begingroup$ I can't wait until someone sees fit to publish anything on Abel-Plana's effect on the MRB constant. If they do, some of your integrals might get posted on its Mathworld page. $\endgroup$ – Marvin Ray Burns Jan 25 at 9:44
  • $\begingroup$ Mathworld used one of your integrals! $\endgroup$ – Marvin Ray Burns Feb 10 at 13:45
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To break the ice, since everyone else is afraid to speak up, here is my answer as to a working model that is a fit the formula for the MRB constant, $$\sum^\infty_{k=1} (-1)^k\left(\sqrt[k]{k}-1\right)=\underset{N\to \infty }{\text{lim}}\left(\sum _{k=1}^{2 N} (-1)^k k^{1/k}\right).$$ If I'm wrong please post a correction! I am passionate to hear your answer and greatly desire to give the bounty to the best one!!!!

First say one was to "save the economy" by borrowing 1.00 dollar at a cost of 100% interest per period, compounded at the end of each period period: The effective rate of interest earned since the beginning wold be $-\sqrt[1]{1}$ times 100%=-100%

Next they loaned 1.00 dollar to someone else and earned 200% interest per 2 periods, compounded at the end of each period: The rate effective of interest earned since the beginning wold be $\sqrt{2}-\sqrt[1]{1}$ all times 100%=41.4214%.

Next, say they were to borrow another dollar at a cost of 300% interest per 3 periods, compounded at the end of each period: The effective rate of interest earned since the beginning wold be $\sqrt{2}-\sqrt[1]{1}-\sqrt[3]{3}$ all times 100%=-102.804%.

Next they loaned yet another to someone else and earned 400% interest per 4 periods, compounded at the end of each period: The effective rate of interest earned since the beginning wold be $\sqrt{2}-\sqrt[1]{1}-\sqrt[3]{3}+\sqrt[4]{4}$ all times 100%=38.6178%.

Again, say they were to borrow another dollar at a cost of 500% interest per 5 periods, compounded at the end of each period: The effective rate of interest earned since the beginning wold be $\sqrt{2}-\sqrt[1]{1}-\sqrt[3]{3}+\sqrt[4]{4}-\sqrt[5]{5}$ all times 100%=-99.3552%.

Say the person and their heirs kept this habit up for many centuries and finally stopped after collecting their last interest payment from a so called someone else . Then the effective rate of interest earned each period since the beginning would be $$\underset{N\to \infty }{\text{lim}}\left(\sum _{k=1}^{2 N} (-1)^k k^{1/k}\right)$$ times 100%=18.7859%. (That is 100% times the MRB constant.)

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