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I came across the following summation, which I would like to estimate. I only need an answer which is correct up to a constant multiple; one can assume that $a, b, c$ are real numbers in the range $[0,1]$ which are small. (The estimate can assume that $a,b,c$ are arbitrarily small): $$ S = \sum_{k \geq 2} \frac{(1+a)^k}{(b^{\frac{1}{k-1}} + c)^k} $$

Note that as long as $a \leq c$ this sum converges, for the denominator approaches to $(1+c)^k$ as $k \rightarrow \infty$.

Really, I only need a bound of the form $$ S = O(f(a,b,c)) $$ for $a,b,c$ sufficiently small.

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Fix $N$ sufficiently large and assuming $a<c$ . We have $$\sum_{k=2}^{N}\left(\frac{1+a}{b^{1/(1-k)}+c}\right)^{k}\leq\sum_{k=2}^{N}\left(\frac{1+a}{b^{1/(1-N)}+c}\right)^{k}.$$ Let $$r:=\frac{1+a}{b^{1/(1-N)}+c};$$ if $N$ is large we have $r<1$ so $$\sum_{k=2}^{N}r^{k}=\frac{r^{2}-r^{N+1}}{1-r}.$$ Now note that $$r^{N+1}\rightarrow0$$ and $$\frac{1+a}{b^{1/(1-N)}+c}\rightarrow\frac{1+a}{1+c}$$ as $N\rightarrow\infty.$ So $$\sum_{k\geq2}\left(\frac{1+a}{b^{1/(1-k)}+c}\right)^{k}=\lim_{N\rightarrow\infty}\sum_{k=2}^{N}\left(\frac{1+a}{b^{1/(1-k)}+c}\right)^{k}\leq\left(\frac{1+a}{1+c}\right)^{2}\left(\frac{1+c}{c-a}\right)=\frac{1+a}{\left(1+c\right)\left(c-a\right)}.$$

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