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:


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.

  • 2
    $\begingroup$ Note on formatting: instead of \big and friends, adding \left( \right) to the delimiters will autoscale them to fit. You can also do \left| stuff \middle/ morestuff \right|, and nest these groupings. For the first equation, you can do \left. (dot means no actual delimiter) to get the slash sized correctly. $\endgroup$ – obscurans May 23 at 2:45
  • $\begingroup$ Oh thanks! Ill update soon. $\endgroup$ – MONODA43 May 23 at 2:46
  • $\begingroup$ Don't worry, I've cleaned it, just needs approval. Once it lands you can see the raw code as-changed. $\endgroup$ – obscurans May 23 at 2:46
  • $\begingroup$ There is a problem in your calculation. $\sum f_i/N$ is not $f_i,$ where in the former the index $i$ is dummy whine in the latter the index $i$ is a certain number. $\endgroup$ – User May 23 at 2:57

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?
| cite | improve this answer | |

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.

| cite | improve this answer | |

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 $.


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 $.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.