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That is, under what conditions would

$$ \sum_{i = 1}^n \frac{a_i}{b_i}= \frac{\sum_{i = 1}^n a_i}{\sum_{i = 1}^n b_i} $$

be true? What about for infinite summations, i.e. when $n \rightarrow \infty$?

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    Rarely. I doubt there is a better condition that guarantees this than "that the two sums are equal".2012-09-14
  • 0
    Why is it that you want to know - is there some context to the question?2012-09-14
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    It would be a more natural question, to my way of thinking, if the left-hand-side (which is the sum of $n$ fractions) were divided by $n$ to match the single fraction on the right-hand-side.2012-09-14
  • 0
    The sum on the right side of the equality is equal to each of the fractions on the left if all of those are equal to each other. But that's a different question.2012-09-14
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    It is actually a subtle diophantine-like equation. It seems very difficult, if not unsolved.2012-09-14
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    It seems clearer to me to write it as $$\sum_j\sum_i \frac{b_i}{b_j}a_j = \sum a_j$$2012-09-14
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    Some additional information is needed. Are the $a_i$, $b_i$ integers? Also, note that $\sum \frac{a_i}{b_i}$ does not chane when $a_1$ and $b_1$ are replaced by $1000a_1$ and $1000b_1$. However, the right-hand side usually changes.2012-09-14
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    I would recommend that before you worry about infinite summations you solve the case $n=2$. What do you get if you start with $(a/b)+(c/d)=(a+c)/(b+d)$?2012-09-17

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The following result might be of help:

Theorem: If $a_i\ge 0$, $b_i>0$ for all $i$ and not all $a_i$s are zero, then $$ \sum_{i = 1}^n \frac{a_i}{b_i}= \frac{\sum_{i = 1}^n a_i}{\sum_{i = 1}^n b_i} $$ does not hold.

Proof: If $a_i\ge 0$, $b_i>0$ for all $i$, we can show by mathematical induction that $$\frac{\sum_{i = 1}^n a_i}{\sum_{i = 1}^n b_i}\le \max_{1\le i\le n}\frac{a_i}{b_i}.\ \ (*)$$

and $$\max_{1\le i\le n}\frac{a_i}{b_i}\le\sum_{i = 1}^n \frac{a_i}{b_i}.\ \ (**)$$

The equality of $(*)$ holds when $a_1/b_1=\cdots=a_n/b_n$, and the equality of $(**)$ holds when at most one of $a_i$s are nonzero; this suggests the equality of $(*)$ and $(**)$ hold at the same time only when all $a_i$s are zero.

Therefore, if $a_i\ge 0$, $b_i>0$ for all $i$ and not all $a_i$s are zero, $$ \sum_{i = 1}^n \frac{a_i}{b_i}>\frac{\sum_{i = 1}^n a_i}{\sum_{i = 1}^n b_i}. $$