# What does this series converge to?

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?

• 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 $S=\sum_{i=1}^\infty s_i$? – celtschk Jul 24 '13 at 15:00

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}
• that doesnt make sense because that means it converges to $n$ right? – Robert Jefferies Jul 24 '13 at 14:48
$$\sum_{i=1}^n\frac{S-s_i}{S}=\sum_{i=1}^n\left(1-\frac{s_i}{S}\right)=n-\frac{\sum_{i=1}^ns_i}{S}=n-\frac SS=n-1$$