Home Competitions IMC 1996 Problem 12
i) Prove that for every sequence ( a n ) n ∈ N (a_{n})_{n\in \mathbb{N}} ( a n ) n ∈ N , such that a n > 0 a_{n}>0 a n > 0 for all n ∈ N n \in \mathbb{N} n ∈ N and ∑ n = 1 ∞ a n < ∞ \sum_{n=1}^{\infty}a_{n}<\infty ∑ n = 1 ∞ a n < ∞ , we have ∑ n = 1 ∞ ( a 1 a 2 ⋯ a n ) 1 n < e ∑ n = 1 ∞ a n . \sum_{n=1}^{\infty}(a_{1}a_{2} \cdots a_{n})^{\frac{1}{n}}< e\sum_{n=1}^{\infty}a_{n}. n = 1 ∑ ∞ ( a 1 a 2 ⋯ a n ) n 1 < e n = 1 ∑ ∞ a n . ii) Prove that for every ϵ > 0 \epsilon>0 ϵ > 0 there exists a sequence ( b n ) n ∈ N (b_{n})_{n\in \mathbb{N}} ( b n ) n ∈ N such that b n > 0 b_{n}>0 b n > 0 for all n ∈ N n \in \mathbb{N} n ∈ N and ∑ n = 1 ∞ b n < ∞ \sum_{n=1}^{\infty}b_{n}<\infty ∑ n = 1 ∞ b n < ∞ and ∑ n = 1 ∞ ( b 1 b 2 ⋯ b n ) 1 n > ( e − ϵ ) ∑ n = 1 ∞ b n . \sum_{n=1}^{\infty}(b_{1}b_{2} \cdots b_{n})^{\frac{1}{n}}> (e-\epsilon)\sum_{n=1}^{\infty}b_{n}. n = 1 ∑ ∞ ( b 1 b 2 ⋯ b n ) n 1 > ( e − ϵ ) n = 1 ∑ ∞ b n .
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