I'm trying to make the sum stop before the summation has a negative exponent.

For example, I would want the sum to stop at $2^0$ in


The sum I'm dealing with is $$\sum P\left(\frac{a}{b}\right)^{\frac{1}{c}(d-fn)}$$ $$\text{Where}\; \{f,a,b,c,d,P\}\in \mathbb{N}$$

The first version I thought of was $$\sum\limits_{n=0}^{?}\frac{1}{2}P\left(\frac{a}{b}\right)^{\frac{1}{c}(d-fn)}(1+\text{sgn}(d-fn))$$ I don't know a good stopping point for this. Technically I could sum to infinity, but that seems unnecessary when the sum could stop after 3 terms. It also fails if $d=fn$ for the final sum as the last sum is divided by 2.

My next version was $$\sum\limits_{n=0}^{\alpha}P\left(\frac{a}{b}\right)^{\frac{1}{c}(d-fn)}$$ $$\alpha = \text{floor}\left(\frac{d}{f}\right)$$ But I don't know if placing $\alpha$ as the upper limit and stating $\alpha = \text{floor}\left(\frac{d}{f}\right)$ is acceptable notation.

Any input on this would be helpful.

$\scriptsize\text{(Hints for a possible closed form would also be appreciated)}$

  • $\begingroup$ If it's the correct terminating value, there's nothing wrong with $$\sum_{n=0}^{\lfloor d/f\rfloor}\cdots$$ $\endgroup$ – MPW Dec 30 '15 at 18:02

The most concise way to express it would be by a summation sign, with the subscripts under it being $n\in\Bbb N$ and $d−fn>0$.

However, there is indeed a closed form. Let $\alpha=d-fm$ be the smallest positive value of $d-fn$.

Then what you are looking for is, $\sum_{n=0}^m P(\frac{a}{b})^{\frac{1}{c}(d-fn)}$, which is the same thing as $$\sum_{n=0}^m P(\frac{a}{b})^{\frac{1}{c}(\alpha+fn)}=\left(P(\frac{a}{b})^{\frac{\alpha}{c}}\right)\sum_{n=0}^m\left(P(\frac{a}{b})^{\frac{f}{c}}\right)^n$$

Then, you would use the geometric series expansion to get the closed form for the above. Namely, if $\beta=P(\frac{a}{b})^{\frac{f}{c}}$, then the above would equal $\left(P(\frac{a}{b})^{\frac{\alpha}{c}}\right)*\frac{\beta^{m+1}-1}{\beta-1}$.


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