\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{Shortlist 2006 A2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 4}
\maketitle

\section*{Problem}
The sequence of real numbers $a_0,a_1,a_2,\dots$ is defined recursively by
\[a_0=-1, \qquad \sum_{k=0}^n \frac{a_{n-k}}{k+1} = 0
  \quad\text{for } n \geq 1.\]
Show that $ a_{n} > 0$ for all $ n\geq 1$.



\section*{External Link}
\url{https://aops.com/community/p821388}

\newpage

\section*{Solution}
In fact, it suffices to take the two consecutive equations
\begin{align*}
  0 &= \frac{a_n}{1} + \frac{a_{n-1}}{2} + \frac{a_{n-2}}{3}
    + \dots + \frac{a_0}{n+1} \\
  0 &= \frac{a_{n+1}}{1} + \frac{a_n}{2} + \frac{a_{n-1}}{3} +
    \frac{a_{n-2}}{4}
    + \dots + \frac{a_0}{n+2}
\end{align*}
and subtract $\frac{n+1}{n+2}$ times the first equation from the second
(eliminating the $a_0$) to obtain a new recursion
\begin{align*}
  a_{n+1} &= \left( \frac{n+1}{n+2} \cdot \frac11 - \frac12 \right) a_n
  + \left( \frac{n+1}{n+2} \cdot \frac12 - \frac13 \right) a_{n-1} \\
  &\qquad+ \left( \frac{n+1}{n+2} \cdot \frac13 - \frac14 \right) a_{n-2} + \dots
  + \left( \frac{n+1}{n+2} \cdot \frac1n - \frac1{n+1} \right) a_1.
\end{align*}
This recursion has the property
that all the coefficients in parentheses are positive.
Since $a_1 = 1/2 > 0$, this implies the result by induction.

\end{document}
