Bounding a "geometric sum in the exponent" 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.
 A: 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.
A: 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.
