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 $n \rightarrow \infty$?

  • 3
    $\begingroup$ Rarely. I doubt there is a better condition that guarantees this than "that the two sums are equal". $\endgroup$ Sep 14, 2012 at 17:23
  • $\begingroup$ Why is it that you want to know - is there some context to the question? $\endgroup$ Sep 14, 2012 at 17:25
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 14, 2012 at 17:28
  • $\begingroup$ 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. $\endgroup$ Sep 14, 2012 at 17:29
  • 2
    $\begingroup$ I would recommend that before you worry about infinite summations you solve the case $n=2$. What do you get if you start with $(a/b)+(c/d)=(a+c)/(b+d)$? $\endgroup$ Sep 17, 2012 at 12:53

1 Answer 1


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


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