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I'm having trouble evaluating this formula. $$\prod_{k=501}^z(1.14+.005(\bigg{\lceil}{\frac{k}{500}}\bigg{\rceil}))$$

I've attempted to simplify this into an algebraic expression which is easier to deal with, but I have no effort to show from it, because the ceiling function switches every $500$ values of $k$, which has thrown off most of my results.

I then attempted to insert this function in to a computational engine like Wolfram Alpha, but every time it is evaluated, it returns $0$. This seems completely counter-intuitive as the sum is always larger than $1$, so the product should always be positive and increasing exponentially; however, I need to be able to simplify this into an algebraic expression or come up with a close model of the function which replicates it's values with accuracy. Any help would be greatly appreciated.

I've tried replacing the $\big{\lceil}\frac{k}{500}\big{\rceil}$ with continous expressions such as $(1+\frac{k}{500})$. However, these results still haven't returned anything useful.

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Replacing $\lceil \frac{k}{500} \rceil$ with $(1 + \frac{k}{500})$ gives you the expression \begin{align*}\prod_{k=501}^{z+500} \Big( 1.14 + 0.005 (1 + k/500) \Big) &= \prod_{k=1}^z \Big( 1.14 + 0.005 k / 500 \Big) \\ &= (0.005/500)^{z} \frac{\Gamma(z + 1.14 + 0.005/500)}{\Gamma(1.14 + 0.005/500)}. \end{align*} Let $\tilde z = z + 1.14001$; then Stirling's approximation gives the upper bound $$\frac{\sqrt{2\pi}}{\Gamma(1.14001)} 10^{-5z} \tilde z^{\tilde z - 1/2} e^{-\tilde z + \frac{1}{12 \tilde z}}.$$

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  • $\begingroup$ I'm a little confused as to how this formula holds. The upper bound goes to zero extremely quickly for values above z=0, but the product (with the ceiling function) should be (1.14+.005)(1.14+.005)...(1.14+.005)(1.14+.01)...(1.14+.01)(1.14+.015).... This value has to remain positive and exponential because you're always multiplying by a number greater than one. Do you have an explanation for why the result is zero? $\endgroup$ – Greyson Cox May 30 '17 at 12:51
  • $\begingroup$ @GreysonCox This bound is not even close to zero. At $z=100$ it's about $1.9 \times 10^{58}$. $\endgroup$ – user420261 May 30 '17 at 14:38
  • $\begingroup$ I know. Which is why every application returning 0, as the function tended to infinity, was so concerning. $\endgroup$ – Greyson Cox Jul 3 '17 at 1:16

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