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This is a cross-post from https://math.stackexchange.com/questions/1174998/estimating-an-unusual-infinite-sum, which didn't get any useful answers

I encountered the following unusual type of infinite sum: $$ S = \sum_{k=1}^{\infty} (a (1 - b^{\frac{1}{k-1}} c))^k$$ where here $a,b,c \in [0,1]$ and $a (1 - c) < 1$. (Interpret the summand when $k = 1$ as simply $a$)

This sum converges, since as $k \rightarrow \infty$ the summand converges to $(a (1-c))^k$. I want to find an upper bound for it. I'm having trouble pinning down its behavior.

An estimate up to a constant multiple would be fine with me.

Thanks for the help!

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  • $\begingroup$ Did you try approximating it by an integral? $\endgroup$
    – Dima Pasechnik
    Commented Mar 9, 2015 at 19:57
  • $\begingroup$ Apparently this does not have a closed form integral. $\endgroup$
    – David Harris
    Commented Mar 9, 2015 at 21:03
  • $\begingroup$ Why not just bound the difficult term between 1 and (1-c)? Are you looking for a really nice function of b to use to sandwich between the two simple bounds? $\endgroup$
    – The Masked Avenger
    Commented Mar 9, 2015 at 22:17
  • $\begingroup$ Alternarively, choose epsilon and find k_0 so that for all larger k, b^{1/k} is within epsilon of 1. Then you are a finite sum away from an upper bound involving (1 - (1- epsilon)c), which might serve your purpose. $\endgroup$
    – The Masked Avenger
    Commented Mar 9, 2015 at 22:23
  • $\begingroup$ In the integral, I'd try to replace the part that causes the absence of a "nice" form by a simpler upper bound, e.g. taking first few terms of the Taylor expansion. $\endgroup$
    – Dima Pasechnik
    Commented Mar 10, 2015 at 8:10

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