limn→∞1−2+3−4+5−6+.......2n√n2+1+√4n2−1is equal to:
-1/3
limn→∞1−2+3−4+5−6+.......2n√n2+1+√4n2−1=limn→∞[1+3+5+7+.....+(2n−1)]−(2+4+6+.....+2n)n√1+1n2+n√4−1n2=limn→∞n2(2.1+(n−1).2)−n2[2.2+(n−1).2]n[√1+1n2+√4−1n2]=limn→∞n2.2n−n22(n+1)n[√1+1n2+√4−1n2]=limn→∞n2−n2−nn[√1+1n2+√4−1n2]=limn→∞−nn[√1+1n2+√4−1n2]=limn→∞−1[√1+1n2+√4−1n2]=−11+2=−13