I am student who learns from examples, and I have yet to see what happens when two sums such as this one,

$$\displaystyle\sum\limits_{a=1}^2 \displaystyle\sum\limits_{b=a+1}^2 4\left[\left(\frac{1}{b-a}\right)^8 - \left(\frac{1}{b-a}\right)^4\right]$$

be solved. I am sure, when a = 1, the inner sum should solve itself like any "normal summation", but when a = 2, what happens? Do I skip the process of the inner summation; do I proceed with the inner sum, and if so, in what ways do I solve the inner sum; or do I solve this problem in a completely different way?

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    $\begingroup$ when the lower index is one greater than the upper index, then even in most general interpretations, the sum is $0$. Therefore, when $a=2$, the inner sum is $0$. $\endgroup$ – robjohn Jun 26 '13 at 23:56
  • $\begingroup$ When would the inner sum not be 0? $\endgroup$ – Flair Jun 27 '13 at 0:20
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    $\begingroup$ I was speaking in more generality. If the upper limit is one less than the lower limit, no terms are included in the sum and that is why the sum is $0$. However, in this particular case, even when $a=1$ and $b=2$, the summand $4\left[\left(\frac1{2-1}\right)^8-\left(\frac1{2-1}\right)^4\right]$ is $0$. $\endgroup$ – robjohn Jun 27 '13 at 0:30

Thinking inductively on the upper and lower indices, we can think of a summation as a difference of two functions on the integers: $$ \sum_{k=a}^bf(k)=F(b)-F(a-1) $$ Even in this generalization, if $b=a-1$, the sum is $0$ irregardless of $f$.


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