IMC 2011 · Day 2 · P8

Define $f(n) = \ln \frac{n+1}{n}$ for $n \ge 1$, and observe that $f(2n) + f(2n+1) = f(n)$. The well-known inequality $\ln(1+x) \le x$ implies $f(n) \le 1/n$. Furthermore introduce $$g(n) = \sum_{k=n}^{2n-1} f^3(k) < n f^3(n) \le 1/n^2.$$ Then $$\begin{align*} g(n) - g(n+1) &= f^3(n) - f^3(2n) - f^3(2n+1) \\ &= (f(2n) + f(2n+1))^3 - f^3(2n) - f^3(2n+1) \\ &= 3 (f(2n) + f(2n+1)) f(2n) f(2n+1) \\ &= 3 f(n) f(2n) f(2n+1), \end{align*}$$ therefore $$\sum_{n=1}^{N} f(n) f(2n) f(2n+1) = \frac{1}{3} \sum_{n=1}^{N} g(n) - g(n+1) = \frac{1}{3} (g(1) - g(N+1)).$$ Since $g(N+1) \to 0$ as $N \to \infty$, the value of the considered sum hence is $$\sum_{n=1}^{\infty} f(n) f(2n) f(2n+1) = \frac{1}{3} g(1) = \frac{1}{3} \ln^3(2).$$