Tight asymptotic bound for recursive algorithm
Suppose that $n = 2^k$. Then
$$ T(n) = T(2^k) = T(2^{k-1}) + 1 = T(2^{k-2}) + 2 = \cdots = T(2^0) + k = k + 1. $$
So $T(n) = \log_2 n + 1$ in that case. In the general case you would get something similar, $T(n) = \lfloor \log_2 n + 1\rfloor$.