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How do you use the direct comparison test to determine if #Sigma 1/(3n^2+2)# from #[1,oo)# is convergent or divergent?

Answer 1

Cameron Alexander

See below.

#3n^2+2 > 3n^2# so

#1/3sum_(k=1)^oo 1/n^2 > sum_(k=1)^oo 1/(3n^2+2)#

We know that #sum_(k=1)^oo 1/n^2# is convergent so #sum_(k=1)^oo 1/(3n^2+2)# also is

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Answer 2

Charles Anctil

To use the direct comparison test to determine if ( \sum_{n=1}^{\infty} \frac{1}{3n^2 + 2} ) from ( n = 1 ) to ( \infty ) is convergent or divergent:

Find a series whose terms are easier to compare with.
Note that for all ( n \geq 1 ), ( 3n^2 + 2 > n^2 ), so ( \frac{1}{3n^2 + 2} < \frac{1}{n^2} ).
Since ( \sum_{n=1}^{\infty} \frac{1}{n^2} ) is a known convergent series (it's a p-series with ( p = 2 > 1 )), we can apply the direct comparison test.
By the direct comparison test, if ( \sum_{n=1}^{\infty} \frac{1}{n^2} ) converges and ( \frac{1}{3n^2 + 2} < \frac{1}{n^2} ) for all ( n \geq 1 ), then ( \sum_{n=1}^{\infty} \frac{1}{3n^2 + 2} ) converges.
Therefore, ( \sum_{n=1}^{\infty} \frac{1}{3n^2 + 2} ) is convergent.

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Answer from HIX Tutor

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.