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

The series is divergent

As,

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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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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.

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.

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.

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