Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n \rightarrow \infty$?

share|improve this question
3  
Rarely. I doubt there is a better condition that guarantees this than "that the two sums are equal". –  Alex Becker Sep 14 '12 at 17:23
    
Why is it that you want to know - is there some context to the question? –  Mark Bennet Sep 14 '12 at 17:25
2  
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. –  Mark Bennet Sep 14 '12 at 17:28
    
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. –  Michael Hardy Sep 14 '12 at 17:29
2  
I would recommend that before you worry about infinite summations you solve the case $n=2$. What do you get if you start with $(a/b)+(c/d)=(a+c)/(b+d)$? –  Gerry Myerson Sep 17 '12 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, $$ \sum_{i = 1}^n \frac{a_i}{b_i}>\frac{\sum_{i = 1}^n a_i}{\sum_{i = 1}^n b_i}. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.