Home > Calculus > Direct Comparison Test for Convergence of an Infinite Series

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

Answer 1

Evelyn Agard

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

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Answer 2

Samantha Adkison

To use the direct comparison test for the series (\sum_{n=2}^{\infty} \frac{1}{\sqrt{n} - 1}), we need to compare it with a known series whose convergence behavior is established. We can choose to compare it with the harmonic series, (\sum_{n=1}^{\infty} \frac{1}{n}), since it is a commonly known divergent series. By comparing the terms of the given series with the terms of the harmonic series, we can determine if the given series is convergent or divergent.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.