Let and both for
If and , then find the minimum and maximum such that
for all
Input your answer as .
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This proof is actually lifted verbatim from my longer proof to Two Looming Towers of Power by Patrick Corn. I rely on this lemma heavily while constructing a proof to that question.
We claim that k = 1 , l = 2 .
To prove that a n + 1 < b n , we proceed with induction.
Base case n = 2 :
1 6 = a 3 < b 2 = 2 7 is obviously true.
Assuming a n + 1 < b n is true, we derive
a n + 1 lo g 2 < a n + 1 lo g 3 < b n lo g 3
lo g 2 a n + 1 < lo g 3 b n
2 a n + 1 < 3 b n
a n + 2 < b n + 1 as required.
To prove that b n < a n + 2 , we proceed by induction again.
We will actually prove a stronger condition, that is 4 b n < a n + 2
Base case n = 2 :
1 0 8 = 4 b 2 < a 4 = 2 1 6 is obviously true.
Assuming 4 b n < a n + 2 is true,
4 b n + 1 = 4 . 3 b n < 4 b n . 3 b n < 1 6 b n = 2 4 b n < 2 a n + 2 = a n + 3 as required.
The second part of the inequality is proven by Deedlit from MathStackExchange . I could never have applied mathematical induction so skilfully myself!