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I have $f_1, f_2, \dots, f_n$ series such $f_1 > f_2 > f_3 > \dots > f_n$.

Also $\sum_{i=1}^n f_i=1$ and all fractions $f_i$ are between $0$ and $1.$

Does there exist a closed form upper bound of the following series.

$$f_1+\frac{f_2}{f_2+f_3+\dots +f_n}+\frac{f_3}{f_3+f_4+\dots +f_n}+\dots +\frac{f_n}{f_n}.$$

A simple upper bound of the series is $n$ since all individual parts are between 0 and 1. Is there a way to simplify the sum of series further preferably into some closed form ?

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    $\begingroup$ I'll just note that if you take a geometric sequence $f_2 = k f_1$, $f_3 = k^2 f_1$, and so on, then at the limit $k \to 0$ the value of the sum goes to $n$, so the upper bound of $n$ is in some sense sharp. $\endgroup$ – Connor Harris Jan 25 '18 at 17:20
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In each of the $n$ terms, numerator is strictly less than denominator.

Hence each term is less than $1$.

Therefore,

$$\color{blue}{\frac{f_1}{1}+\frac{f_2}{f_2+f_3+\dots +f_n}+\frac{f_3}{f_3+f_4+\dots +f_n}+\dots +\frac{f_n}{f_n} <1+1+ \ldots + 1=n}$$

LHS will tend $ \to n$, when $$f_1 >> f_2>>f_3 \cdots >>f_n \quad ; \sum f_i = 1$$

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  • $\begingroup$ Thanks for this upper bound but it seems very loose. I understand that $n$ upper bound exists. I am looking for a tighter upper bound if possible. $\endgroup$ – learner Jan 25 '18 at 17:35
  • $\begingroup$ @learner This bound can be achieved as close as you want, so it's not at all loose. $\endgroup$ – Jaideep Khare Jan 25 '18 at 17:41

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