Bounding a "geometric sum in the exponent"

Question

I'm interested in an upper bound for the following quantity: $$\sum_{i = 1}^n c^{(3/4)^i},\quad c\in (0,1)$$

This initially looks like a geometric sum, but has the property that the "$r^i$" part is in the exponent. One can get a trivial upper bound on this quantity via the bound $c \leq 1$, which gives: $$\sum_{i = 1}^n c^{(3/4)^i} \leq n$$ This isn't good enough for my purposes. What would be good enough is any bound of the form: $$\sum_{i = 1}^n c^{(3/4)^i} <n$$ In short, I'm interested in any bound that's non-trivial. Does anyone know of any work on this problem (or more genrally upper bounding any sums of the form $\sum_i \exp(r^i)$)? Part of my difficulty is in trying to find the right thing to search, as "upper bound exponential sum" leads to questions in analytic number theory where the exponential is $\exp(2\pi if(x) / q)$, which is quite different.

Answer 1

The continuous version of your sum can be expressed by composition of Ei function and exponential function, where the Ei function asymptotic approximation can be found here.

Answer 2

Throughout this, let $r = 3/4$. This answer will requires that: $$\int c^{r^x}\mathsf{d}x = \frac{\mathsf{Ei}(r^x\ln(c))}{\ln(1/r)} + C$$ and the bounds: $$\frac{1}{2}\exp(-x)\ln(1 + (2/x)) \leq E_1(x) \leq \exp(-x) \ln(1 + (1/x))$$ for $x \geq 0$, where: $$E_1(x) = -\mathsf{Ei}(-x)$$ Since $i\mapsto c^{r^i}$ is increasing, we have that: $$\sum_{i = 1}^n c^{r^i} \leq \int_{i = 1}^{n+1}c^{r^x}\mathsf{d}x = \frac{\mathsf{Ei}(r^{n+1}\ln(c))-\mathsf{Ei}(r\ln(c))}{\ln(1/r)}$$ Now, as $\ln(c) < 0$ (because $c\in(0,1)$) and $r^k\geq 0$, we have that the arguments of $\mathsf{Ei}$ are negative. For this reason, we can write: $$\mathsf{Ei}(r^k\ln(c)) = - E_1(-r^k\ln(c)) = E_1(r^k\ln(1/c))$$ We then have: $$\sum_{i = 1}^n c^{r^i} \leq \frac{E_1(r\ln(1/c)) - E_1(r^{n+1}\ln(1/c))}{\ln(1/r)}$$ Now, the aformentioned upper bound on $E_1(x)$ gives: $$E_1(r\ln(1/c)) \leq \exp(-r\ln(1/c))\ln\left(1 + \frac{1}{r\ln(1/c)}\right) = c^r\ln\left(1 + \frac{1}{r\ln(1/c)}\right)$$ Moreover, the lower bounds give: $$-E_1(r^{n+1}\ln(1/c)) \leq \frac{1}{2}\exp(-r^{n+1}\ln(1/c))\ln\left(1 + \frac{2}{r^{n+1}\ln(1/c)}\right) = \frac{1}{2}c^{r^{n+1}}\ln\left(1 + \frac{2}{r^{n+1}\ln(1/c)}\right)$$ These combine to give: $$\sum_{i = 1}^{n}c^{r^i}\leq\frac{1}{\ln(1/r)}\left(c^r\ln\left(1 + \frac{1}{r\ln(1/c)}\right) - \frac{1}{2}c^{r^{n+1}}\ln\left(1 + \frac{2}{r^{n+1}\ln(1/c)}\right)\right)$$ It's likely that this can be simplified further (i.e. via bounds on $\log(1+x)$), but from here the problem has been reduced to a more "standard" question, via an (experimentally) fairly tight bound.