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

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To use the direct comparison test for the series ( \sum \frac{1}{n-1} ) from ( n = 1 ) to infinity, we compare it with the series ( \sum \frac{1}{n} ). Since for all ( n \geq 2 ), ( \frac{1}{n-1} ) is less than or equal to ( \frac{1}{n} ), and ( \sum \frac{1}{n} ) is the harmonic series which diverges, we conclude that ( \sum \frac{1}{n-1} ) also diverges.

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To use the direct comparison test to determine if the series ( \sum_{n=1}^{\infty} \frac{1}{n-1} ) is convergent or divergent, we need to compare it to a known series whose convergence or divergence is already established.

In this case, we can compare the given series to the harmonic series, ( \sum_{n=1}^{\infty} \frac{1}{n} ), which is a well-known divergent series.

To apply the direct comparison test, we need to show that the given series is either less than or greater than the harmonic series for all ( n ) beyond some point.

Consider ( n = 2 ) and beyond:

For ( n \geq 2 ), we have ( \frac{1}{n-1} \geq \frac{1}{n} ), because ( n-1 ) is always less than or equal to ( n ).

Since ( \frac{1}{n-1} \geq \frac{1}{n} ) for all ( n \geq 2 ), and the harmonic series ( \sum_{n=1}^{\infty} \frac{1}{n} ) is known to be divergent, by the direct comparison test, the series ( \sum_{n=1}^{\infty} \frac{1}{n-1} ) is also 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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