How do you test the series #Sigma (n^n)/(lnn)^n# from #n=[2,oo)# by the ratio test?
It's really easier in this case to use the root test:
So that:
Thus the series is not convergent, and as it has positive terms, it is divergent:
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To test the series Σ(n^n)/(ln(n))^n from n=2 to infinity using the ratio test:

Calculate the ratio of consecutive terms: ( \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}/(ln(n+1))^{n+1}}{n^n/(ln(n))^n} ).

Simplify the ratio: ( \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{n^n} \times \frac{(ln(n))^n}{(ln(n+1))^{n+1}} ).

Take the limit as n approaches infinity of the absolute value of this ratio.

If the limit is less than 1, the series converges. If the limit is greater than 1 or it doesn't exist, the series diverges.
So, the series Σ(n^n)/(ln(n))^n converges if ( \lim_{{n \to \infty}} \left \frac{(n+1)^{n+1}}{n^n} \times \frac{(ln(n))^n}{(ln(n+1))^{n+1}} \right < 1 ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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