Difference between $\sum_i\frac{a_i}{b_i}$ and $\frac{\sum_i a_i}{\sum_i b_i}$ Take ratio $\frac{a}{b}$ such that $a>0$, $b>0$. The $a$ can be some measurement and $b$ reference, for example. Having more of such measurements, arithmetic mean of these ratios is
$$\frac{1}{N}\sum\limits_{i=1}^{N}\frac{a_i}{b_i}$$
Another expression that intuitively feels like average is
$$\frac{\sum\limits_{i=1}^N a_i}{\sum\limits_{i=1}^N b_i}$$
Am I just confused and the second formula has no practical significance, or it relates to the first one in some way? They are not the same, but is it possible they converge to the same value for $N\to\infty$?
The ratio $\frac{a_i}{b_i}$ for represents savings. Take $a_i$ algorithm run time after optimization and $b_i$ algorithm run time before optimization, where $b_i\geq a_i$. Then $\sum_{i=1}^{\infty}a_i \to \infty$, $\sum_{i=1}^{\infty}b_i \to \infty$ and $$\frac{\sum_{i=1}^{\infty}a_i}{\sum_{i=1}^{\infty}b_i} \in [0,1]$$
The motivation is as follows: from standpoint where arithmetic mean cannot be used directly I hope to approximate it using the second formula. It seems to give similar results in practice, but I need formal validation why this can be used instead (if it can, of course). 
 A: Perhaps this could help..
If you write out the sums you will get.
$$\sum_{i=1}^{n}\frac{a_i}{b_i}=\frac{a_1}{b_1}+\frac{a_2}{b_2}+\frac{a_3}{b_3}+\cdots+\frac{a_n}{b_n}$$
And
 $$\frac{\sum_{i=1}^{n}{a_i}}{\sum_{i=1}^{n}b_i}=\frac{a_1+a_2+a_3+\cdots+a_n}{b_1+b_2+b_3+\cdots+b_n}$$ 
As you can see you have to find sequences where..
$$\frac{a_1+a_2+a_3+\cdots+a_n}{b_1+b_2+b_3+\cdots+b_n}=\frac{a_1}{b_1}+\frac{a_2}{b_2}+\frac{a_3}{b_3}+\cdots+\frac{a_n}{b_n}$$
Now suppose $\sum_{i=1}^\infty a_i$ and $\sum_{i=1}^\infty b_i$ converge. That does not mean $\sum_{i=1}^\infty \frac{a_i}{b_i}$ will converge. Just substitute $a_i={\frac{1}{2}}^i$ and $b_i=\frac{1}{3}^{i}$. You'll find that
$\sum_{i=1}^{\infty}\frac{\frac{1}{2^i}}{\frac{1}{3^i}}\neq\frac{\sum_{i=1}^{\infty}\frac{1}{2^i}}{\sum_{i=1}^{\infty}\frac{1}{3^i}}$.
Rarely would converging sequences in the form of  be the same. So I do not see much potential with comparing $\sum_{i=1}^{n}\frac{a_i}{b_i}$ with $\frac{\sum_{i=1}^{n}{a_i}}{\sum_{i=1}^{n}b_i}$. 

EDIT:
If both $\sum_{i=0}^{\infty}a_i$ and $\sum_{i=0}^{\infty}b_i$ diverges but the sum of $\sum_{i=0}^{\infty}\frac{a_i}{b_i}$ converges then in most cases $\frac{\sum_{i=0}^{\infty}a_i}{\sum_{i=0}^{\infty}b_i}$ and $\sum_{i=0}^{\infty}\frac{a_i}{b_i}$ will not have the same value. While $\sum_{i=0}^{\infty}\frac{a_i}{b_i}$ will mostly converge to a number greater than zero , $\frac{\sum_{i=0}^{\infty}a_i}{\sum_{i=0}^{\infty}b_i}$ will mostly converge to zero(if $\lim_{i\to\infty}\frac{a_i}{b_i}=0$) or $\infty$ ($\lim_{i\to\infty}\frac{a_i}{b_i}=\infty$).
However if your using the aritmetic mean, where $$\lim_{i\to\infty}\frac{a_i}{b_i}=c$$ then yes. $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^n \frac{a_i}{b_i}=\lim_{n\to\infty}\frac{\sum_{i=0}^n {a_i}}{\sum_{i=0}^{n}b_i}$$
This is simple to prove because if $\lim_{i\to\infty} \frac{a_i}{b_i}=c$ then we get $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}c$ with partial sum formula as $\lim_{n\to\infty}\frac{1}{n}(cn)=c$. As for $\lim_{n\to\infty}\frac{\sum_{i=0}^{n}{a_i}}{\sum_{i=0}^n b_i}$ we should treat this as the limit of a function divided by another function. If $\lim_{i\to\infty}\frac{a_i}{b_i}=c$ then the poportion is the use $\lim_{i\to\infty}a_i=c\lim_{i\to\infty}(b_i)$ and thus $\frac{\sum_{i=0}^{\infty}a_i}{\sum_{i=0}^{\infty}b_i}=\frac{c\sum_{i=0}^{\infty}(b_i)}{\sum_{i=0}^{\infty}b_i}=c$ 
In this case you can use this identity when $a_i$ and $b_i$ nearly the same or have their limit ratio approach zero. Identities or theorems should make mathematical problems simple. At best, sum indentities on both sides of the equation apply to broad types of functions, not just specfic instances. This is similar to comparing $\lim_{x\to\infty}\frac{\ln(x)}{x}$ to $\lim_{x\to\infty}\frac{x^2}{2^x}$. They both converge to the same limit but comparing them both will not make solving either easier
A: If $a_i$ is the number of miles traveled during the $i$th leg of a journey and $b_i$ is the number of hours that leg took, then $\displaystyle \frac{\sum_{i=1}^N a_i}{\sum_{i=1}^N b_i}$ is the average speed in miles per hour.
