If n=(1999)!, then 1999∑x=1lognx is equal to
[AMU 1999; DCE 2005]
limπ→∞∑nk=1 kn2+k2 is equals to [Roorkee 1999]
limn→∞[1n+1n+1+1n+2+⋯+12n]= [Karnataka CET 1999]