For all `n in N, 1 + 1/(sqrt(2))+1/(sqrt(3))+1/(sqrt(4))++1/(sqrt(n))`

Statement-1: For every natural number nge2 , (1)/(sqrt(1))+(1)/(sqrt(2))+(1)/(sqrt(3))+...+(1)/(sqrt(n))gtsqrt(n) Statement-2: For every natural number nge2, sqrt(n(n+1))ltn+1

lim_ (n rarr oo) sum_ (n = 1) ^ (n) (sqrt (n)) / (sqrt (r) (3sqrt (r) + 4sqrt (n)) ^ (2))

lim_(n rarr oo)(sin(1)/(sqrt((n))))((1)/(sqrt(n+1)))^(+(1)/(sqrt(n+2))+(1)/(sqrt(n+2)))

If a_1,a_2,a_3,………a_n are in A.P, where a_igt0 for all i show that 1/(sqrt(a_1)+sqrt(a_2))+1/(sqrt(a_2)+sqrt(a_3))+……..+1/(sqrt(a_(n-1))+sqrt(a_n))=(n-1)/(sqrt(a_1)+sqrt(a_n))

If a_(1),a_(2),a_(3),a_(n) are in A.P,where a_(i)>0 for all i, show that (1)/(sqrt(a_(1))+sqrt(a_(2)))+(1)/(sqrt(a_(2))+sqrt(a_(3)))++(1)/(sqrt(a_(n-1))+sqrt(a_(n)))=(n-1)/(sqrt(a_(1))+sqrt(a_(n)))

Use induction to show that for all n in N . sqrt(a+sqrt(a+sqrt(a+....+sqrt(a))))lt (1+sqrt((4a+1)))/(2) where'a' is fixed positive number and n radical signs are taken on LHS.