# How do you apply the ratio test to determine if #Sigma 1/(ln(lnn))^n# from #n=[3,oo)# is convergent to divergent?

The ratio test states that:

Therefore, using the ratio test for the example above gives:

This is true since any number between 0 and 1, when raised to any positive power, must always still be between 0 and 1.

Back to the original problem:

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To apply the ratio test to determine if (\sum_{n=3}^{\infty} \frac{1}{(\ln(\ln n))^n}) is convergent or divergent, we compute the limit:

[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|]

where (a_n = \frac{1}{(\ln(\ln n))^n}).

Compute (a_{n+1}) and (a_n), then take their ratio. If the limit is less than 1, the series is convergent; if it's greater than 1 or diverges to infinity, the series is divergent; if it's exactly 1, the ratio test is inconclusive.

After taking the limit, you'll need to analyze whether it's less than, greater than, or equal to 1.

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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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