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 $(\sum f_ia_i)/(\sum a_i)\to f$

Question

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.

Answer 1

I see a few problems with the proof:

1. 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)
2. 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.
3. 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?

Answer 2

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.

Answer 3

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 $.