Let a1=1a_1 = 1a1=1, an=1n∑k=1n−1akan−ka_n = \dfrac{1}{n} \sum\limits_{k=1}^{n-1} a_k a_{n-k}an=n1k=1∑n−1akan−k for n≥2n \ge 2n≥2. Show that
(i) lim supn→∞∣an∣1/n<2−1/2\limsup\limits_{n \to \infty} |a_n|^{1/n} < 2^{-1/2}n→∞limsup∣an∣1/n<2−1/2;
(ii) lim supn→∞∣an∣1/n≥2/3\limsup\limits_{n \to \infty} |a_n|^{1/n} \ge 2/3n→∞limsup∣an∣1/n≥2/3.
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