Let a 1 be the maximum of f ( x ) = 1/ ( 2 + ⌊ sin x ⌋) . For n ≥ 1 , set a n + 1 = a n + ( − 1 ) n + 2 / ( n + 1 ) . Find lim n → ∞ a n .

Since ⌊ sin x ⌋ ∈ { − 1 , 0 , 1 } , a 1 = 1 . The recurrence gives a n = 1 − 1/2 + 1/3 − ⋯ + ( − 1 ) n + 1 / n , tending to ln 2 .

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Let $a_{1}$ be the maximum of $f (x) = 1/ (2 + ⌊ sin x ⌋)$ . For $n \geq 1$ , set $a_{n + 1} = a_{n} + (- 1)^{n + 2} / (n + 1)$ . Find $lim_{n \to \infty} a_{n}$ .

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Since $⌊ sin x ⌋ \in {- 1, 0, 1}, a_{1} = 1$ . The recurrence gives $a_{n} = 1 - 1/2 + 1/3 - \dots + (- 1)^{n + 1} / n$ , tending to $ln 2$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

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