# How do you apply the ratio test to determine if #Sigma 1/(lnn)^n# from #n=[2,oo)# is convergent to divergent?

he series:

is convergent.

We have the series:

Now evaluate the ratio:

Now consider the function:

and therefore:

Then:

which proves the series to be convergent.

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To apply the ratio test to determine the convergence or divergence of the series Σ 1/(ln n)^n from n=2 to infinity, you calculate the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term. If the limit is less than 1, the series is convergent. If the limit is greater than 1 or does not exist, the series is divergent.

The (n+1)th term of the series is 1/(ln(n+1))^(n+1). Therefore, the ratio of the (n+1)th term to the nth term is [1/(ln(n+1))^(n+1)] / [1/(ln n)^n]. Simplifying this gives [(ln n)^n] / [(ln(n+1))^(n+1)]. Taking the limit as n approaches infinity of this ratio will determine the convergence or divergence of the series.

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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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