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What does the following expression converge to?

$${\sum_{i = 1}^n{\left(\frac{S-s_i}{S}\right)^S}}$$

Where the sum of the $s_i$'s equals $S$.

How do you work out what it converges to?

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    $\begingroup$ Which series? As is, you've got a finite sum. Do you mean $\lim_{n\to\infty}$ of that sum? Also, is $S=\sum_{i=1}^n s_i$ or $S=\sum_{i=1}^\infty s_i$? $\endgroup$ – celtschk Jul 24 '13 at 15:00
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EDIT2: Assuming that you have $$ \sum_{i=1}^{\infty} s_i = S $$ and you want to determine the convergence of $$ \sum_{i=1}^{\infty}\left(\frac{S - s_i}{S}\right)^S $$ note first that $$ \lim_{i\to \infty} s_i = 0. $$ Then $$ \lim_{i\to \infty}\left(\frac{S - s_i}{S}\right)^S \neq 0. $$ So the series is not convergent.

EDIT: This is the answer to the question as originally stated: You have $$\begin{align} \sum_{i=1}^n \frac{S - s_i}{S} &= \frac{1}{S}\sum_{i=1}^n S - s_i \\&= \frac{1}{S}\left[nS - \sum_{i=1}^n s_i\right] \\ &= n - \frac{1}{S}\sum_{i=1}^n s_i. \end{align} $$

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  • $\begingroup$ that doesnt make sense because that means it converges to $n$ right? $\endgroup$ – Robert Jefferies Jul 24 '13 at 14:48
  • $\begingroup$ oh wait I forgot a crutial part please excuse $\endgroup$ – Robert Jefferies Jul 24 '13 at 14:49
  • $\begingroup$ please look at my edit $\endgroup$ – Robert Jefferies Jul 24 '13 at 14:50
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$$ \sum_{i=1}^n\frac{S-s_i}{S}=\sum_{i=1}^n\left(1-\frac{s_i}{S}\right)=n-\frac{\sum_{i=1}^ns_i}{S}=n-\frac SS=n-1$$

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  • $\begingroup$ I apologize please look at my edit $\endgroup$ – Robert Jefferies Jul 24 '13 at 14:50

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