Calculate the sum,
$$e^{ar}+e^{ar^2}+...+e^{ar^n}\ \text{where} \ a,r\in \mathbb{R}$$
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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Arithmetic progression, geometric progression – Aug 25 '19 at 14:36
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1See https://en.wikipedia.org/wiki/Lacunary_function – lhf Aug 25 '19 at 15:01
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2Note that $e^{r^2} \ne (e^r)^2$ !!!! – Ted Shifrin Aug 25 '19 at 17:04
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1@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. – graveolensa Aug 27 '19 at 22:52
1 Answers
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Let ${ar}^i=i\log m.$ Then the series becomes $$\sum_{i=1}^{n}{e^{ar^{i}}}=\sum_{i=1}^{n}{e^{i\log{m}}}=\sum_{i=1}^{n}{{m}^i},$$ which is a geometric series.
Allawonder
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