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!
Best Answer
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 $z\in\mathbb{C}$.