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 $n \rightarrow \infty$?
Tell me more
×
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.
|
|
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, $$ \sum_{i = 1}^n \frac{a_i}{b_i}>\frac{\sum_{i = 1}^n a_i}{\sum_{i = 1}^n b_i}. $$ |
||||
|
|
