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.

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  1. August 3, 2016 at 4:55 am

    I did the very same thing with difference in the fact that i used able’s summation by parts formula at the start . Eventually i was lead to prove something very similar to last of inequalities of this post.

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