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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!