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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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    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

1 Answers 1

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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}. $