Calculate the sum,
$$e^{ar}+e^{ar^2}+\dots+e^{ar^n}\ \text{where} \ a,r\in \mathbb{R}$$
It's easy to calculate the sum when the powers of $e$ are in an arithmetic progression. How do we proceed when the powers are in geometric progression?
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$\begingroup$ Arithmetic progression, geometric progression $\endgroup$user694028– user6940282019-08-25 14:36:22 +00:00Commented Aug 25, 2019 at 14:36
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1$\begingroup$ See en.wikipedia.org/wiki/Lacunary_function $\endgroup$lhf– lhf2019-08-25 15:01:59 +00:00Commented Aug 25, 2019 at 15:01
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3$\begingroup$ Note that $e^{r^2} \ne (e^r)^2$ !!!! $\endgroup$Ted Shifrin– Ted Shifrin2019-08-25 17:04:34 +00:00Commented Aug 25, 2019 at 17:04
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3$\begingroup$ @lhf: this is finite, and should not be regarded as a lacunary function -- the functional equation does not apply, and for a fair number of $a$ and $r$ it does not converge when carried out to infinity. $\endgroup$graveolensa– graveolensa2019-08-27 22:52:19 +00:00Commented Aug 27, 2019 at 22:52
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2$\begingroup$ It is essentially a duplicate of this. $\endgroup$ultralegend5385– ultralegend53852025-02-26 08:21:12 +00:00Commented Feb 26, 2025 at 8:21
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1 Answer
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This is what we call a Lacunary Exponential Sum, which yields no simple general form, and specifically if $r\ne 1$. There are special cases in which the sums do simplify, which are for $r=1$ and $r=0$, which you can solve straightforwardly. What we can do, however, is analyze this via Taylor-Expansion. Behold
$\begin{align} \text{Let }S_n&=e^{ar}+e^{ar^2}+e^{ar^3}+\dots+e^{ar^n}\\ e^{ar^{k}}&=\displaystyle\sum_{j=0}^{\infty}\frac{(ar^k)^j}{j!}\\ S_n&=\sum_{k=1}^{n}\sum_{j=0}^{\infty}\frac{a^j(r^j)^k}{j!} &&\text{Substituting Back Into The Total Sum $S_n$}\\ S_n&=\sum_{j=0}^{\infty}\frac{a^j}{j!}\Bigg(\sum_{k=1}^{n}(r^j)^{k}\Bigg)\\ \sum_{k=1}^{n}(r^j)^{k}&=\frac{r^j(1-r^{jn})}{1-r^j} &&\text{Standard Geometric Progression}\\ \text{Thus, $S_n$}&=\sum_{j=0}^{\infty}\frac{a^j}{j!}\bigg(\frac{r^j(1-r^{jn})}{1-r^j}\bigg) \end{align}$
A note on the algebra done here. We switched the order of summation, provided that $S_n$ converges, which the inner summation does for fixed values of $k$. It is not the most elegant of solutions, but at least one way to look at it.
