limn→∞ 1+2+3+…+n1+3+5+…+(2n−1)=
1
3/2
12
2
We have,
limn→∞ 1+2+3+…+n1+3+5+…+(2n−1)=limn→∞ n(n+1)2n2=limn→∞ n(n+1)2n2=limn→∞ n2+n2n2=12