The sequence would be:
n + n/2 + n/3 + n/4 + n/5 + ...+ n/n
= n . [1 + 1/2 + 1/3 + 1/4 + ...+ 1/n ]
Lets denote the summation part as S.
So, S
= 1 + 1/2 + 1/3 + 1/4 + ...+ 1/n
= 1 + (1/2 + 1/3) + (1/4 + 1/5 + 1/6 + 1/7) + (1/8 + 1/9 + ...+ 1/15) + ...+ ( 1/(n/2) + ... + n)
= 1 + (1/2 + 1/2) + (1/4+ 1/4+ 1/4 + 1/4) + (1/8 + 1/8 + ...+ 1/8) + ...+ ( 1/(n/2) + ... + 1/(n/2) )
Now, let x=1/2
So, S
= 1+ (2x) + (4x2) + (8x3) + ...+ (n/2).(x)log(n/2)
= (2x)0 + (2x)1 + (2x)2 + ... + (2x)log(n/2)
Now, 2x = 2 * (1/2) = 1
$\therefore$ S = 1 + 1 + 1 + ...( log(n/2) times ) = logn
Hence, T(n) = $\theta(n.logn)$