ru sure it should be sloved usin indef inte coz it can be solved by usin limit of a sum...

 

let n=1000,

     

_{=[r=0 ]}^{[r=1000 ]} 1/n+r

 

_{=[r=0 ]}^{[ r=1000]} (1/n)/(1+r/n)

let r/n=x,sigma gets converted into integral

 

_{=[0 ]}^{[1 ]} 1/1+x

 

=[log(1+x)]₀¹

=log2