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What if on the sum there is a fraction in the limit?

$\sum_{m=k/12}^{k}$ or $\sum_{m=0}^{k/12+1}$

thank you very much!

what type of sequence is used for summing this type of interval?

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Where (and in what context) did you encounter such a creature? It will be easier for folks to answer your questions if you provide as much relevant information as possible. For example, if we happened to know that $k$ is some integer divisible by $12$, this wouldn't be at all problematic. –  Cameron Buie Jun 26 '12 at 19:09
    
@user1434040: I just wanted to point out that you've flagged your own question as "not constructive". Is that what you intended to do? –  Zev Chonoles Jun 27 '12 at 17:38
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1 Answer 1

up vote 2 down vote accepted

In the first case, the sum is over all integers $m$ that are in the closed interval whose endpoints are $\frac{k}{12}$ and $k$.

In the second case, the sum is over all integers $m$ that are in the closed interval whose endpoints are $0$ and $\frac{k}{12} +1$.

(The convoluted language is because in principle $k$ could be negative.)

Remarks: $1$. Part of the convention is that (unless something is said to the contrary) an index called $m$ ranges over integers.

The languages of mathematics, like other human languages, have many such conventions. Often they are not made explicit.

$2$. Probably in principle one should use the appropriate floor or ceiling functions, but they can sometimes interfere with quick understanding, particularly in a subscript or superscript.

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it is possible to change the limits? –  nanme Jun 26 '12 at 19:12
    
@user1434040: Sure. But I thought that your two examples were "generic" enough to illustrate the general pattern. One could say that if $a$ and $b$ are reals, then the ordinary meaning of $\sum_{m=a}^b$ is the sum over all integers $m$ in a certain interval. –  André Nicolas Jun 26 '12 at 19:19
    
but how can this be evaluated? thank you very much! –  nanme Jun 26 '12 at 19:32
    
@user1434040: Let $k=30$. Then for example $\sum_{m=k/12}^k m^2=3^2+4^2+\cdots+30^2$. –  André Nicolas Jun 26 '12 at 19:40
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