How do you use the comparison test for #sum (((ln n)^3) / (n^2))# n=1 to #n=oo#?
You've written:
The Limit Comparison Test says (paraphrased):
Therefore, let:
So, now we get:
(Nono, don't use L'Hopital's Rule here; there's an easy way to do this.)
This limit can be split into a product of limits, like so:
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To use the comparison test for the series ( \sum_{n=1}^\infty \frac{(\ln n)^3}{n^2} ):

We need to find a series ( \sum_{n=1}^\infty b_n ) such that ( 0 \leq b_n \leq a_n ), where ( a_n = \frac{(\ln n)^3}{n^2} ) is the given series.

We choose ( b_n ) such that it's easier to evaluate and compare to ( a_n ).

Since ( (\ln n)^3 ) grows slower than ( n^2 ) as ( n ) approaches infinity, we can choose ( b_n = \frac{1}{n^2} ).

Now, we compare ( a_n ) and ( b_n ): ( \frac{(\ln n)^3}{n^2} \leq \frac{1}{n^2} ).

Since ( \frac{(\ln n)^3}{n^2} ) is always less than or equal to ( \frac{1}{n^2} ) for ( n \geq 1 ), and ( \sum_{n=1}^\infty \frac{1}{n^2} ) converges (it's a pseries with ( p = 2 > 1 )), by the comparison test, ( \sum_{n=1}^\infty \frac{(\ln n)^3}{n^2} ) also converges.
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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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