# How to approximate this large sum of exponential terms

Is there any way to approximate the following sum:

$$\sum\limits_{i_1=1}^N\sum\limits_{i_1=2}^N \cdots \sum\limits_{i_k=1}^N \cdots\sum\limits_{i_N=1}^N \exp(-r_{i_1}-r_{i_{k+1}}-r_{i_{2k+1}}- r_{i_{3k+1}}\cdots -r_{i_N})$$

where $k<<N$, $i_1 \neq i_2 \neq i_k \cdots \neq i_N$ and $r_{i_j}$ is any positive variable. Final expression could be in terms of $r_{i_j}$. Please explain and thanks.

• What is $r_{i_{j}}$ related to? Also what is the need for $k << N$ if there are $N$ sums? – Leucippus Jul 5 '14 at 20:56
• Here $r_{i_j}$ is any positive variable. I put $k<<N$ in case we could approximate for large N. – upol94 Jul 5 '14 at 20:59

Since \begin{align} \sum_{k=1}^{m} x^{k} = -1 + \frac{1-x^{m+1}}{1-x} = \frac{x(1-x^{m})}{1-x} \end{align} then \begin{align} \sum_{k=1}^{m} e^{-k} = \frac{1-e^{-m}}{e-1}. \end{align} Each sum that does not have an index of the summed function provides the value $N$. As the summation appears \begin{align} \sum_{i_1=1}^{N} \sum_{i_2=1}^{N} \cdots \sum_{i_k=1}^{N} \cdots\sum_{i_N=1}^{N} e^{-i_1-i_{k+1}-i_{2k+1}- \cdots - i_{N}} \end{align} the result is \begin{align} N^{2k} \left( \frac{1-e^{-N}}{e-1} \right)^{N-2k}. \end{align} If the summation is supposed to read \begin{align} \sum_{i_1=1}^{N} \sum_{i_2=1}^{N} \cdots \sum_{i_k=1}^{N} \cdots\sum_{i_N=1}^{N} e^{-i_1 - \cdots - i_{k} - \cdots - i_{N}} \end{align} then the value is \begin{align} \left( \frac{1-e^{-N}}{e-1} \right)^{N}. \end{align}
• Thanks a lot for the answer. I misspelled the variables $r$. Can you please update the answers and why there is an $m$ term in the final expression ? – upol94 Jul 5 '14 at 20:32
• The new notation is still ambiguous. Can you explicitly write the equation in the case $N=2$ i.e.: 8 terms without $\Sigma$ – JJacquelin Jul 6 '14 at 6:06
It is a little difficult to understand that the whole sum is the product to $N$ times the simple sum $\sum\limits_{i=1}^N \exp(-i)$. So, I prefer to propose a recursive process :
• The Sum does not include all the terms. Only $r_{i_1}+r_{i_{k+1}}+r_{i_{2k+1}}+ \cdots$ – upol94 Jul 5 '14 at 20:49
• I found the question is still ambiguous with its new wording. Would you mind write explicitly (without $\Sigma$) the equation in a particular case, for example $N=2$ , i.e.: the 4 terms of the sum. – JJacquelin Jul 6 '14 at 8:58
• There is somthing fishy. It seems that your example doesn't corresponds to the initial wording : $\sum\limits_{i_1=1}^N\sum\limits_{i_1=2}^N \cdots \sum\limits_{i_k=1}^N \cdots\sum\limits_{i_N=1}^N \exp(-r_{i_1}-r_{i_{k+1}}-r_{i_{2k+1}}- r_{i_{3k+1}}\cdots -r_{i_N})$ In case $N=3$, the sum includes $3^3=27$ terms. If some are not taken into account, it should be specified what is the rule to eliminate them from the sum. – JJacquelin Jul 6 '14 at 14:31
• OK, I see. But I was confused by your first wording : $\sum\limits_{i_1=1}^N\sum\limits_{i_1=2}^N \cdots \sum\limits_{i_k=1}^N \cdots\sum\limits_{i_N=1}^N \exp(-r_{i_1}-r_{i_{k+1}}-r_{i_{2k+1}}- r_{i_{3k+1}}\cdots -r_{i_N})$ where $i_1=1$ and $i_k=1$ and $i_N=1$ : they are not different. You should have written $i_k=k$ since all the index $1, 2, ..., (k-1)$ were used before. And it remains only $i_N=N$ for the last one. I hope someone else will continue to help you. Best regards. – JJacquelin Jul 6 '14 at 15:00