TITLE: Merge Sort time complexity analysis
QUESTION [5 upvotes]: How can I prove that $T(n) = 2T(n/2) + n$ is $O(n \log n)$ ?

REPLY [4 votes]: T(1) = 1 
T(n) = 2T (n/2) + cn 
T(n) = 2T (n/2) + cn
T(n/2) = 2T(n/4) + c (n/2)
T(n) = 2 [ 2T (n/4) + c (n/2) ] + cn
T(n) = 4T (n/4) + 2cn
Similary, 
T(n) = 8T (n/8) + 3cn 
General 
T(n) = 2^k T (n/2^K) + kcn
(n/2^k) = 1
n = 2^k
k = log base2 n
plug in k into the general formula
T(n) = n T(1) + logBase2n cn 
T(n) = Big Thetha(n log n )