C=limn→∞(1+12+13+⋯+1n−logn)=limx→∞(∑n≤x1n−logx).
∑n≤x1n=logx+1−∫1∞t−[t]t2dt⏟This is C as shown below+O(1x).
∑n≤x1n−logx=1−∫1∞t−[t]t2dt+O(1x).
C=limx→∞(∑n≤x1n−logx)=limx→∞(1−∫1∞t−[t]t2dt+O(1x))=1−∫1∞t−[t]t2dt.