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

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To determine the convergence or divergence of the series ( \sum_{n=2}^{\infty} \frac{\ln n}{n+1} ) using the direct comparison test, we compare it to a known series.

We notice that ( \frac{\ln n}{n+1} ) is similar to ( \frac{1}{n} ) as ( n ) approaches infinity, since the natural logarithm grows slower than any power of ( n ).

Now, we compare the given series to the harmonic series ( \sum_{n=1}^{\infty} \frac{1}{n} ), which is known to be divergent. Specifically, we compare ( \frac{\ln n}{n+1} ) to ( \frac{1}{n} ) for ( n \geq 2 ).

Since ( \frac{\ln n}{n+1} < \frac{1}{n} ) for all ( n \geq 2 ), and the harmonic series ( \sum_{n=1}^{\infty} \frac{1}{n} ) diverges, by the direct comparison test, the series ( \sum_{n=2}^{\infty} \frac{\ln n}{n+1} ) also diverges.

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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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