What's a good class of functions for bounding/comparing ratios of complicated logarithms? I have th

fabios3 2022-06-25 Answered
What's a good class of functions for bounding/comparing ratios of complicated logarithms?
I have this goofy series n = 2 log 2 [ n log 2 2 n ] n log 2 2 n that Wolfram Alpha tells me diverges by the comparison test (and indeed, in the larger problem I'm working on I need to prove that the expression containing it diverges), but I'm struggling to find a good divergent lower bound.
I can't just throw away all the inner logarithms--then I run into the theorem lim x log a x x b 0 b > 0
So I'm looking at things like log 2 [ n 3 ] n log 2 2 n (larger, unfortunately), log 2 [ n 3 ] n 2 log 2 n (converges), and log 2 [ n 2 ] n log 2 2 n (still too large).
Is there a more general class or form I can use to find a simple divergent lower bound, instead of stabbing in the dark?
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Answers (2)

Jaida Sanders
Answered 2022-06-26 Author has 18 answers
Logarithms are really small, compared to any polynomial (or any positive power of n). So one very frequently useful trick is to try to ignore the logarithm, since it's not going to have a huge affect on the convergence of the series. More precisely, we have that
n log 2 2 n n
n log 2 2 n n
by a little bit, so let's just test the series with n, rather than the complicated argument inside the logarithm. This leads to
n log 2 [ n log 2 2 n ] n log 2 2 n n log 2 n n log 2 2 n = n 1 n log 2 n =
Another thing that suggests this approach is to rewrite the numerator as
log 2 n + log 2 ( log 2 2 n ) = log 2 n + 2 log 2 log 2 n
If log 2 n is small relative to n, then log 2 log 2 n is tiny in comparison. This strongly suggests comparison to the series without this term.

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Abram Boyd
Answered 2022-06-27 Author has 5 answers
Lemma n = 2 log 2 n n diverges.
Proof By comparison to n = 2 1 n
For n 2, log 2 n 1. So log 2 n n 1 n . Since n = 2 1 n diverges, n = 2 log 2 n n diverges.

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