How do you use the direct comparison test to determine if #Sigma 1/(sqrtn-1)# from #[2,oo)# is convergent or divergent?

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Answer 1 · 2017-12-06T18:22:14

The series is divergent

Let #f(n)=1/(sqrtn-1)#

On the interval #[2, oo)#, the function #f(n)# is positive , continuous and decreasing.

Let #a_n=1/(sqrtn-1)#

#sqrtn-1<=sqrtn#

#1/(sqrtn-1)>=1/sqrtn#

#b_n=1/sqrtn#

#a_n>=b_n#

#lim_(p->oo)int_2^p1/sqrtxdx=lim_(p->oo)[2sqrtx]_2^p#

#=lim_(p->oo)(2sqrtp-2sqrt2)=+oo#

As,

#b_n# diverges, so #a_n# diverges by the comparison test