How do you use the limit comparison test to determine if #Sigma (2n^21)/(3n^5+2n+1)# from #[1,oo)# is convergent or divergent?
The series:
is convergent.
Given:
We can now consider that:
and
so that:
is convergent.
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To use the limit comparison test on the series ( \sum \frac{2n^21}{3n^5+2n+1} ) from ( n = 1 ) to ( \infty ):

Select a simpler series that you know the convergence of.

Compute the limit: [ \lim_{{n \to \infty}} \frac{{\frac{{2n^2  1}}{{3n^5 + 2n + 1}}}}{{\frac{1}{{n^p}}}} ] where ( p ) is the degree of the highest power in the denominator of the original series.

If the limit is a finite positive number, then both series either converge or diverge. If it's zero or infinite, the behavior of both series may differ.

Choose a simpler series wisely. Common choices include ( \sum \frac{1}{n^p} ) or ( \sum \frac{1}{n} ) depending on the degree of the polynomials in the original series.

Simplify the limit as much as possible.
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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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