a_n =(−1)^n (⌊(n/9)⌋+1) Σ_(n≥0) a_n =? b_n =(((a_n −1)(9−a_n ))/(n+1)) Σ_(n≥0) b


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Question Number 2329 by 123456 last updated on 16/Nov/15

$a_{n} = {(- 1)}^{n} (⌊ \frac{n}{9} ⌋ + 1)$ $\sum_{n ⩾ 0} a_{n} = ?$ $b_{n} = \frac{(a_{n} - 1) (9 - a_{n})}{n + 1}$ $\sum_{n ⩾ 0} b_{n} = ?$
Commented by prakash jain last updated on 16/Nov/15

$a_{0} = 1, a_{1} = - 1, a_{2} = 1 . ., a_{8} = 1, \underset{n = 0}{\sum^{8}} a_{n} = 1$ $a_{9} = 2, a_{10} = - 2, a_{11} = 2 . ., a_{17} = 2, \underset{n = 9}{\sum^{n = 17}} a_{n} = 2$ $⌊ \frac{n}{9} ⌋ = k$ $n - k = m$ $\underset{i = 0}{\sum^{n}} a_{i} = \frac{k (k + 1)}{2} + {(- 1)}^{m} \frac{1 + k}{2} + \frac{1 + k}{2}$
Commented by Rasheed Soomro last updated on 16/Nov/15

$⌊ \frac{n}{9} ⌋ b r a c k e t f u n c t i o n ?$
Commented by 123456 last updated on 16/Nov/15

$greatest lower integer$ $⌊ 0, 3 ⌋ = 0$ $⌊ 0, 7 ⌋ = 0$
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