IMC 1996 · Day 2 · P12

(i) Put for $n \in \mathbb{N}$ $$\tag{1} c_n = (n+1)^n / n^{n-1}.$$ Observe that $c_1 c_2 \cdots c_n = (n+1)^n$. Hence, for $n \in \mathbb{N}$, $$(a_1 a_2 \cdots a_n)^{1/n} = (a_1 c_1 a_2 c_2 \cdots a_n c_n)^{1/n} / (n+1) \le (a_1 c_1 + \cdots + a_n c_n) / n(n+1).$$ Consequently, $$\tag{2} \sum_{n=1}^{\infty} (a_1 a_2 \cdots a_n)^{1/n} \le \sum_{n=1}^{\infty} a_n c_n \left( \sum_{m=n}^{\infty} (m(m+1))^{-1} \right).$$ Since $$\sum_{m=n}^{\infty} (m(m+1))^{-1} = \sum_{m=n}^{\infty} \left( \frac{1}{m} - \frac{1}{m+1} \right) = 1/n$$ we have $$\sum_{n=1}^{\infty} a_n c_n \left( \sum_{m=n}^{\infty} (m(m+1))^{-1} \right) = \sum_{n=1}^{\infty} a_n c_n / n = \sum_{n=1}^{\infty} a_n ((n+1)/n)^n < e \sum_{n=1}^{\infty} a_n$$ (by (1)). Combining the last inequality with (2) we get the result.

(ii) Set $a_n = n^{n-1} (n+1)^{-n}$ for $n = 1, 2, \dots, N$ and $a_n = 2^{-n}$ for $n > N$, where $N$ will be chosen later. Then $$\tag{3} (a_1 \cdots a_n)^{1/n} = \frac{1}{n+1}$$ for $n \le N$. Let $K = K(\varepsilon)$ be such that $$\tag{4} \left( \frac{n+1}{n} \right)^n > e - \frac{\varepsilon}{2} \quad \text{for } n > K.$$ Choose $N$ from the condition $$\tag{5} \sum_{n=1}^{K} a_n + \sum_{n=N+1}^{\infty} 2^{-n} \le \frac{\varepsilon}{(2e - \varepsilon)(e - \varepsilon)} \sum_{n=K+1}^{N} \frac{1}{n},$$ which is always possible because the harmonic series diverges. Using (3), (4) and (5) we have $$\begin{align*} \sum_{n=1}^{\infty} a_n &= \sum_{n=1}^{K} a_n + \sum_{n=N+1}^{\infty} 2^{-n} + \sum_{n=K+1}^{N} \frac{1}{n} \left( \frac{n}{n+1} \right)^n < \\ &< \frac{\varepsilon}{(2e - \varepsilon)(e - \varepsilon)} \sum_{n=K+1}^{N} \frac{1}{n} + \left( e - \frac{\varepsilon}{2} \right)^{-1} \sum_{n=K+1}^{N} \frac{1}{n} = \\ &= \frac{1}{e - \varepsilon} \sum_{n=K+1}^{N} \frac{1}{n} \le \frac{1}{e - \varepsilon} \sum_{n=1}^{\infty} (a_1 \cdots a_n)^{1/n}. \end{align*}$$