IMC 2012 · Day 2 · P7

Observe that $$k a_k = \frac{(1 + (k+1) a_k) a_{k+1}}{a_k} = \frac{a_{k+1}}{a_k} + (k+1) a_{k+1} \quad \text{for all } k \ge 1,$$ and hence $$\tag{1} \sum_{k=0}^{n} \frac{a_{k+1}}{a_k} = \frac{a_1}{a_0} + \sum_{k=1}^{n} \bigl( k a_k - (k+1) a_{k+1} \bigr) = \frac{1}{2} + 1 \cdot a_1 - (n+1) a_{n+1} = 1 - (n+1) a_{n+1}$$ for all $n \ge 0$.

By (1) we have $\sum\limits_{k=0}^{n} \frac{a_{k+1}}{a_k} < 1$. Since all terms are positive, this implies that the series $\sum\limits_{k=0}^{\infty} \frac{a_{k+1}}{a_k}$ is convergent. The sequence of terms, $\frac{a_{k+1}}{a_k}$ must converge to zero. In particular, there is an index $n_0$ such that $\frac{a_{k+1}}{a_k} < \frac{1}{2}$ for $n \ge n_0$. Then, by induction on $n$, we have $a_n < \frac{C}{2^n}$ with some positive constant $C$. From $n a_n < \frac{Cn}{2^n}$ we get $n a_n \to 0$, and therefore $$\sum_{k=0}^{\infty} \frac{a_{k+1}}{a_k} = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{a_{k+1}}{a_k} = \lim_{n \to \infty} \bigl( 1 - (n+1) a_{n+1} \bigr) = 1.$$ Remark. The inequality $a_n \le \frac{1}{2^n}$ can be proved by a direct induction as well.