hi ! for a_0 = 1 and ∀ n ≥ 1, a_n = (1/n) Σ_(k=0) ^(n−1) (a_k /(n−k)) . prove that ∀ n ≥ 0, we get 0 ≤ a


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Question Number 138383 by henderson last updated on 12/Apr/21

$hi!$ $for a_{0} = 1 and \forall n ⩾ 1, a_{n} = \frac{1}{n} \underset{k = 0}{\sum^{n - 1}} \frac{a_{k}}{n - k} .$ $prove that \forall n ⩾ 0, we get 0 ⩽ a_{n} ⩽ 1 .$
Commented by mitica last updated on 13/Apr/21

$i n d u c t i e$ $0 ⩽ a_{n} ⩽ 1 \Rightarrow 0 ⩽ a_{n + 1} ⩽ 1$ $a_{n + 1} = \frac{1}{n + 1} \underset{k = 0}{\sum^{n}} \frac{a_{k}}{n + 1 - k} ⩾ 0$ $∣ a_{n + 1} ∣= \frac{1}{n + 1} ∣ \underset{k = 0}{\sum^{n}} \frac{a_{k}}{n + 1 - k} ∣⩽ \frac{1}{n + 1} \underset{k = 0}{\sum^{n}} \frac{∣ a_{k} ∣}{n + 1 - k} ⩽$ $\frac{1}{n + 1} \underset{k = 0}{\sum^{n}} \frac{1}{n + 1 - k} ⩽ \frac{1}{n + 1} \cdot (n + 1) = 1$
Commented by henderson last updated on 13/Apr/21

$thank u, sir mitica!$
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