limπ→∞ 1ρ+2ρ+3ρ+⋯+nρnρ+1= [AIEEE 2002]
1p−1
1p+2
limπ→∞ 1ρ+2ρ+3ρ+⋯+nρnρ+1=limπ→∞∑nr=1[rρnρ+1] =limπ→∞1n∑nr=1(rn)ρ=∫10 xρdx=[xρ+1p+1]10=1p+1
limπ→∞∑nr=1 1nern is [AIEEE 2004]