Using the Arithmetic Mean-Geometric Mean Inequality to prove convergence of a series I am studying for my qualifying exam in analysis and ran across this problem from an old exam:
Let $0<x_n<1$ be an infinite sequence of real numbers such that for all $0<r<1$,
$$\sum_{x_{n}<r}\ln\left(\frac{r}{x_n}\right)\leq 1.$$ Show that $\sum_{n=1}^{\infty}(1-x_n)<\infty$.
I was given a hint to use AM-GM inequality by turning the sum of logs into the log of the products. When I tried this I got the following (using N for the partial sum and product, as infinite AM-GM seems weird), but am not sure where to go from there:
$$1\geq \ln\left(\prod_{x_{n}<r}^{N}r\right)-\ln\left(\prod_{x_{n}<r}^{N}x_n\right)\geq  \ln\left(\prod_{x_{n}<r}^{N}r\right)-\ln\left(\left(\frac{\sum_{x_{n}<r}^{N}x_n}{N}\right)^N\right)$$
It seems like if you exponentiate everything then maybe you can bound the sum by $e$?
Would appreciate any help.
 A: For each $r \in (0,1)$, let $C_r = \{x_n : x_n\lt r\}$. Then $C_r$ is finite and
$$1 \ge \sum_{x\in C_r}\log\left(\frac{r}{x}\right) = \log\prod_{x\in C_r}\frac{r}{x}$$
Therefore
$$\prod_{x\in C_r}x\ge \frac{r^{|C_r|}}{e}$$
By the AM-GM inequality:
$$\frac{1}{|C_r|}\sum_{x\in C_r} x\ge \left(\prod_{x\in C_r}x\right)^{1/|C_r|} \ge r\cdot e^{-1/|C_r|}$$
or:
$$\sum_{x\in C_r} x \ge r\cdot |C_r|\cdot e^{-1/|C_r|}$$
Multiplying by $-1$ and adding $|C_r|$ to both sides this becomes:
$$\sum_{x\in C_r} (1-x) \le  |C_r|\cdot(1 - r\cdot  e^{-1/|C_r|})$$
As $r\to 1$ the left hand side approaches $\sum(1-x_n)$, so the conclusion will follow from the fact (proved next) that the right hand side is bounded.
We have, for any $s,r \in (0,1)$:
$$ 1\ge \sum_{x_n\lt s} \log\left(\frac{s}{x_n}\right) \ge \sum_{x_n\lt rs} \log\left(\frac{s}{x_n}\right)$$
$$\ge \sum_{x_n\lt rs} \log\left(\frac{s}{rs}\right) = |C_{rs}|\log\left(\frac{1}{r}\right)$$
Hence
$$(1-r)|C_{rs}|\le \frac{1-r}{\log(1/r)}\le 1$$
Since this holds for any $s\in(0,1)$ and $\lim_{s\to 1}|C_{rs}| = |C_r|$, we have $(1-r)|C_r|\le 1$ for all $r\in(0,1)$
Now,
$$|C_r|\cdot(1 - r\cdot  e^{-1/|C_r|}) = |C_r|\cdot(1 - r) + r\cdot|C_r|\cdot(1-  e^{-1/|C_r|})$$
$$ \le 1 + |C_r|\cdot(1-  e^{-1/|C_r|}) \le 2$$
Where the last step follows from the fact that $x(1-e^{-1/x})\lt 1$ for all $x\gt 0$.
