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