循环的复杂度 T(n)=T(n−2)+1/lgn？

Question

This is a problem from the CLRS book. The Introduction to Algorithms Study Group website gave the following answer to it:

( http://clrs.skanev.com/04/problems/03.html )

Is this answer right? I don't understand the last two lines.

Answer 1

No, it's not. Also there's a typo, instead of infinity there should be n. For strict mathematical proof you should ask on another stackExchange site (mathematical one). But for your intuition I can show the following.

Let's imagine that n = 2^2^k then sum of 1/lg(i) equals

1/lg2 + 1/lg3 + 1/lg4 + 1/lg5 + 1/lg6 + 1/lg7 + 1/lg8 + 1/lg9 +
1/lg10 + 1/lg11 + 1/lg12 + 1/lg13 + 1/lg14 + 1/lg15 + ... + 1/lg n-1

and this is approximately

1/lg2 + 1/lg2 + 1/lg4 + 1/lg4 + 1/lg4 + 1/lg4 + 1/lg8 + 1/lg8 +
1/lg8 + 1/lg8 + 1/lg8 + 1/lg8 + 1/lg8 + 1/lg8 + ... + 1/lg n-1

equals

1/1 + 1/1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/3 + 1/3 +
1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + ... + 1/ (2^k - 1) (as lg n = 2^k)

After combining we have

sum(1/i * 2^i) from 1 to 2^k-1

where the last member is n/2 / 2^k-1 which is something about 2^(2^kk-1) and this is far not theta of lg lg n = k . And of course the whole sum is even bigger.