Home > Analysis, Inequalities, Olympiad, Problem Solving > IMC 2016 – Day 2 – Problem 6

## IMC 2016 – Day 2 – Problem 6

Problem 6. Let ${(x_1,x_2,...)}$ be a sequence of positive real numbers satisfying ${\displaystyle \sum_{n=1}^\infty \frac{x_n}{2n-1}=1}$. Prove that

$\displaystyle \sum_{k=1}^\infty \sum_{n=1}^k \frac{x_n}{k^2} \leq 2.$

Sketch of proof: We can write the sum to be bounded as

$\displaystyle \sum_{k=1}^\infty (x_1+...+x_k)/k^2.$

Rearranging terms (we can do it since the series is absolutely convergent) we note that we have

$\displaystyle x_1 \sum_{n=1}^\infty \frac{1}{n^2}+x_2\sum_{n=2}^\infty \frac{1}{n^2}+...+x_k\sum_{n=k}^\infty \frac{1}{n^2}+...$

Taking a look at the hypothesis we note that it is enough to prove that

$\displaystyle \sum_{k=n}^\infty \frac{1}{k^2} \leq \frac{2}{2n-1}.$

Since this doesn’t work by induction right away (try it…) let’s try and prove something stronger:

$\displaystyle \sum_{k=n}^\infty \frac{1}{k^2} \leq \frac{1}{n}.$

Since ${1/n-1/n^2 \leq 1/(n+1)}$ the induction will work.