Exercise 3.4

Exercise 4: Find the upper and lower limits of the sequence { s n } defined by

s 1 = 0 , s 2 m = s 2 m − 1 2 , s 2 m + 1 = 1 2 + s 2 m .

Answers

We observe that if a n = ( 2 n − 1 ) ∕ 2 n and b n = ( 2 n − 1 − 1 ) ∕ 2 n , then s 2 n + 1 = a n and s 2 n = b n fulfills both the initial condition and the recursion, so that this is a closed form of the sequence. Note that a n → 1 , b n → 1 2 , and that any subsequence of s n contains either a subsequence of a n or a subsequence of b n . Consequently, any convergent subsequence converges to either 1 or 1 2 . We get

limsup ⁡ n → ∞ s n = 1 liminf ⁡ n → ∞ s n = 1 2
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2023-08-07 00:00
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{ s n } has a lower limit of 0, and an upper limit of 1.

Proof. By definition above, we can see that { s n } = { 0 , 0 , 1 ∕ 2 , 1 ∕ 4 , 3 ∕ 4 , 3 ∕ 8 , … } . We can then define this sequence discretely as two subsequences:

s 2 m − 1 = 1 − 1 2 m − 1 s 2 m = 1 2 − 1 2 m ,

where m ≥ 1 . Each of these subsequences is monotonic and nondecreasing, since the amount we subtract from 1 and 1 ∕ 2 respectively both tend to 0 as m → ∞ . We can then analyze the limits of each subsequence separately. For s 2 m − 1 , we know its limit is its value as m → ∞ , which equals 1. For s 2 m , we know its limit is its value as m → ∞ , which equals 1 2 . Thus, we know that for the original sequence s n , its lower limit is 1 2 (since any other subsequence of s n that differs from { s 2 m } and has infinite number of terms, has terms greater than 1 2 as n → ∞ ) and its upper limit is 1 (since any other subsequence of s n that differs from s 2 m − 1 and has infinite number of terms, has terms less than 1 as n → ∞ ). □

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2023-09-01 19:41
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