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Trying to determine asymptotic of

$$T(n) = T(n-2) + \displaystyle\frac{1}{ \lg n}$$ $$\lg n = \log_{2}n $$

Last term $\frac{1}{ \lg n}$ give me a lot of trouble. Iterative method doesn't work. Tried change of variables, there is nothing to change.

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marked as duplicate by Daniel Fischer Feb 21 '15 at 21:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What is the $\lg$ function? Natural logarithm? – Rod Carvalho Sep 12 '12 at 4:25
@RodCarvalho It is log base 2 (binary log) – newprint Sep 12 '12 at 4:33
@newprint You should accept the answer that you find most useful. You have not accepted answers to any of your questions so far, even though they seem like good answers. – Paresh Feb 9 '13 at 17:17
up vote 1 down vote accepted

Hint: You can bound $T(n)$ between $\frac 12 \sum_{i=4}^n \frac 1{\lg i}$ and $\frac 12 \sum_{i=4}^n \frac 1{\lg (i-2)}$ (to within a small correction), each of which is close to the logarithmic integral $\int \frac 1{\ln x}\; dx$, which goes as $\frac x{ \ln x}$ (like yours to within factors of $\ln 2$)

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Might want to check those small values of $i$ in your sums... – Erick Wong Sep 12 '12 at 4:48

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