Using AM-GM to show that if $\{a_i\},\{f_i\}$ are positive sequences s.t. $\sum a_i=\infty$ and $f_i\to f>0$, then $(\sum f_ia_i)/(\sum a_i)\to f$ This is from DJH Garling's book, Inequalities: A Journey into Linear Analysis
Suppose $\left\{a_i\right\}$ and $\left\{f_i\right\}$ are positive sequences such that:
$$\sum^\infty_{i=1}a_i=\infty$$
and 
$$f_i\rightarrow f>0$$
Show that as $N\rightarrow\infty$:
$$\left.\left(\sum^N_{i=1}f_ia_i\right)\middle/\left(\sum^N_{i=1}a_i\right)\right.\rightarrow f$$
The problem appears in the section on AM-GM, so I assume it should be used somewhere in the proof.
The approaches I've tried so far don't involve AM-GM:
Consider 
$$\left|\left(\sum^N_{i=1}f_ia_i\right)\middle/\left(\sum^N_{i=1}a_i\right)-f\right|=\left|\left(\sum^N_{i=1}f_ia_i\right)\middle/\left(\sum^N_{i=1}a_i\right)-f_i+f_i-f\right|$$
by triangle inequality and convergence of $f_n$ to $f$:
$$\begin{align*}
&\leq\left|\left(\sum^N_{i=1}f_ia_i\right)\middle/\left(\sum^N_{i=1}a_i\right)-f_i\right|+o(1)\\
&=\left|\sum^N_{i=1}f_i\left(\frac{a_i}{\sum^N_{i=1}a_i}-\frac{1}{N}\right)\right| + o(1)
\end{align*}$$
which must be $o(1)$ since $f_n$ converges to a finite value.
Any help seeing where AM-GM could play a role would be much appreciated. I could also use some feedback on what I've tried so far.
 A: I see a few problems with the proof:


*

*Note that $i$ in the sums is precisely what you're summing over (from $1$ to $N$), so $f_i$ makes no sense outside it (in the first line)

*So when you moved $f_i$ into the sum, it's not the same $f_i$ for different terms in the sum, which is invalid.

*I agree $f_i$ converges, but how does that mean the weighted average of $f_i$ under the bizarre weighting
$$\frac{a_i}{\sum_{i=1}^N a_i}-\frac{1}{N}$$
converges to zero, especially as the upper limit of summation $N$ goes to infinity?

A: I provide my opinion, though still without the AM-GM inequality.
The task is to estimate
\begin{align}
\left|\frac {\sum\limits_{i=1}^n f_i a_i}{\sum\limits_{i=1}^n a_i} - f\right|
& = \left| \frac {\sum\limits_{i=1}^n(f-f_i)a_i}{\sum\limits_{i=1}^n a_i} \right|\le \frac {\sum\limits_{i=1}^n|f-f_i|a_i}{\sum\limits_{i=1}^n a_i}.
\end{align}
Given any $\varepsilon>0,$ since $f_i$ tend to $f$ as $i\to\infty,$ we could take $N_1\in\mathbb N$ such that $|f-f_i|< \frac \varepsilon 2$ for all $i\ge N_1.$ Then we can take a larger $N_2\in\mathbb N$ such that
$$
\frac {\sum\limits_{i=1}^{N_1}|f-f_i|a_i}{\sum\limits_{i=1}^{N_2}a_i}<\frac \varepsilon 2
$$
since $\sum\limits_{i=1}^\infty a_i = \infty.$ As a result, we have
\begin{align}
\frac {\sum\limits_{i=1}^n|f-f_i|a_i}{\sum\limits_{i=1}^n a_i}
& \le \frac {\sum\limits_{i=1}^{N_1}|f-f_i|a_i}{\sum\limits_{i=1}^{n}a_i} + \frac {\sum\limits_{i=N_1+1}^{N_2}|f-f_i|a_i}{\sum\limits_{i=1}^{n}a_i}\\
& \le \frac \varepsilon 2 + \frac {\sum\limits_{i=N_1+1}^{N_2}\frac \varepsilon 2 \cdot a_i}{\sum\limits_{i=1}^{n}a_i}\\
& \le \frac \varepsilon 2 + \frac \varepsilon 2 = \varepsilon
\end{align}
for $n\ge N_2,$ and the result follows.
A: I don't see the need
for AM-GM either.
Here's a reasonably rigorous proof
using the usual
bad part/good part method.
If
$a_i > 0, f_i > 0$,
$\sum^\infty_{i=1}a_i=\infty
$
and
$f_i\rightarrow f>0
$
then
show that
$\dfrac{\sum^N_{i=1}f_ia_i}{\sum^N_{i=1}a_i}\to f
$.
Proof.
For any $c > 0$,
there is a $n(c)$ such that
$|f_i-f| < c
$
for
$i > n(c)
$.
Similarly,
since
$ \sum^N_{i=1}a_i \to \infty
$,
for any $r > 0$
there is a
$m(r)$ such that
$ \sum^{m(r)}_{i=1}a_i
\gt r
$.
Then,
for any $N > n(c)$,
$\begin{array}\\
d(N, c)
&=\dfrac{\sum^N_{i=1}f_ia_i}{\sum^N_{i=1}a_i}- f\\
&=\dfrac{\sum^N_{i=1}(f_ia_i-fa_i)}{\sum^N_{i=1}a_i}\\
&=\dfrac{\sum^N_{i=1}a_i(f_i-f)}{\sum^N_{i=1}a_i}\\
&=\dfrac{\sum^{n(c)}_{i=1}a_i(f_i-f)+\sum^N_{i=n(c)+1}a_i(f_i-f)}{\sum^N_{i=1}a_i}
\qquad\text{bad part/good part}\\
&=\dfrac{\sum^{n(c)}_{i=1}a_i(f_i-f)}{\sum^N_{i=1}a_i}+\dfrac{\sum^N_{i=n(c)+1}a_i(f_i-f)}{\sum^N_{i=1}a_i}\\
&=d_1(N, c)+d_2(N, c)\\
d_1(N, c)
&=\dfrac{\sum^{n(c)}_{i=1}a_i(f_i-f)}{\sum^N_{i=1}a_i}\\
|d_1(N, c)|
&=\left|\dfrac{\sum^{n(c)}_{i=1}a_i(f_i-f)}{\sum^N_{i=1}a_i}\right|\\
&\lt c
\qquad\text{for } N > m(\left|\sum^{n(c)}_{i=1}a_i(f_i-f)\right|/c)\\
d_2(N, c)
&=\dfrac{\sum^N_{i=n(c)+1}a_i(f_i-f)}{\sum^N_{i=1}a_i}\\
|d_2(N, c)|
&=\left|\dfrac{\sum^N_{i=n(c)+1}a_i(f_i-f)}{\sum^N_{i=1}a_i}\right|\\
&=\dfrac{\left|\sum^N_{i=n(c)+1}a_i(f_i-f)\right|}{\sum^N_{i=1}a_i}\\
&\le\dfrac{\sum^N_{i=n(c)+1}\left|a_i(f_i-f)\right|}{\sum^N_{i=1}a_i}\\
&\le\dfrac{\sum^N_{i=n(c)+1}\left|a_ic\right|}{\sum^N_{i=1}a_i}
\qquad\text{since } N > n(c)\\
&\le\dfrac{c\sum^N_{i=n(c)+1}\left|a_i\right|}{\sum^N_{i=1}a_i}\\
&\le c
\qquad\text{since } a_i > 0\\
\text{so that}
&\text{ for any } c > 0,
\text{ if } N > \max(n(c), 
 m(\left|\sum^{n(c)}_{i=1}a_i(f_i-f)\right|/c))\\
|d(N, c|
&=|d_1(N, c)+d_2(N, c)|\\
&\le|d_1(N, c)|+|d_2(N, c)|\\
&\lt 2c\\
\end{array}
$
Therefore
$\lim_{N \to \infty} \dfrac{\sum^N_{i=1}f_ia_i}{\sum^N_{i=1}a_i}
-f
=0
$
so
$\lim_{N \to \infty} \dfrac{\sum^N_{i=1}f_ia_i}{\sum^N_{i=1}a_i}
=f
$.
