limn→∞ 1+24+34+…+n4n5−limn→∞ 13+23+33+…+n3n5, is
15
130
0
14
we have,
limn→∞ 14+24+…+n4n5−limn→∞ 13+23+…+n3n5=limn→∞ n(n+1)(2n+1)3n2+3n−130n5−limn→∞ n2(n+1)24n5=limn→∞ 1301+1n2+1n3+3n−1n2−14limn→∞ 1n1+1n2=630−0=15