# How do you show that the series #ln1+ln2+ln3+...+lnn+...# diverges?

We can immediately rewrite as

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By the (nth-term) divergence test:

So the series diverges

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To show that the series (\ln 1 + \ln 2 + \ln 3 + \ldots + \ln n + \ldots) diverges, we can use the comparison test with the harmonic series.

First, notice that (\ln n > 0) for all (n \geq 2), since the natural logarithm is only defined for positive numbers.

Next, we compare the terms of the given series with the terms of the harmonic series, which is known to diverge: [\ln n > \frac{1}{n}] This comparison holds for all (n \geq 2), as (\ln n) grows slower than (\frac{1}{n}).

Now, we can apply the comparison test: since the harmonic series diverges, and we have shown that (\ln n > \frac{1}{n}) for all (n \geq 2), by the comparison test, the series (\ln 1 + \ln 2 + \ln 3 + \ldots + \ln n + \ldots) 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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