Putnam 2011, Problem A-2

Let $a_1, a_2, \ldots$ and $b_1, b_2, \ldots$ be sequences of positive real numbers such that $a_1 = b_1 = 1$ and $b_n = b_{n-1}a_n - 2$ for $n = 2, 3, \ldots$. Assume that the sequence $(b_j)$ is bounded. Prove that
$$S = \sum_{n=1}^{\infty} \frac{1}{a_1 \cdots a_n}$$
converges, and evaluate $S$.

Solution:
For \(m \geq 1\), write  
\[
S_m = \frac{3}{2} \left( 1 - \frac{b_1 \cdots b_m}{(b_1 + 2) \cdots (b_m + 2)} \right).
\]  
Then \(S_1 = 1 = 1/a_1\) and a quick calculation yields  
\[
S_m - S_{m-1} = \frac{b_1 \cdots b_{m-1}}{(b_2 + 2) \cdots (b_m + 2)} = \frac{1}{a_1 \cdots a_m}
\]  
for \(m \geq 2\), since \(a_j = (b_j + 2)/b_{j-1}\) for \(j \geq 2\). It follows that \(S_m = \sum_{n=1}^m 1/(a_1 \cdots a_n)\).  
Now if \((b_j)\) is bounded above by \(B\), then \(\frac{b_j}{b_j + 2} \leq \frac{B}{B + 2}\) for all \(j\), and so \(3/2 > S_m \geq 3/2(1 - (B/(B + 2))^m)\). Since \(B/(B + 2) < 1\), it follows that the sequence \((S_m)\) converges to \(S = 3/2\).