# How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo5/(2n^2+4n+3)# ?

By making the denominator smaller,

Since

converges by Comparison Test.

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To use the Direct Comparison Test on the infinite series ( \sum_{n=1}^{\infty} \frac{5}{2n^2 + 4n + 3} ), we compare it with another series whose convergence behavior is known.

- We need to find a series ( \sum_{n=1}^{\infty} b_n ) such that ( 0 \leq \frac{5}{2n^2 + 4n + 3} \leq b_n ) for all ( n ).
- If ( \sum_{n=1}^{\infty} b_n ) converges, and ( \frac{5}{2n^2 + 4n + 3} ) is bounded above by ( b_n ), then ( \sum_{n=1}^{\infty} \frac{5}{2n^2 + 4n + 3} ) also converges.
- If ( \sum_{n=1}^{\infty} b_n ) diverges, and ( \frac{5}{2n^2 + 4n + 3} ) is bounded below by ( b_n ), then ( \sum_{n=1}^{\infty} \frac{5}{2n^2 + 4n + 3} ) also diverges.

In this case, we notice that ( 2n^2 + 4n + 3 ) is always positive for ( n \geq 1 ). Thus, ( \frac{5}{2n^2 + 4n + 3} ) is always positive.

We can try to bound ( \frac{5}{2n^2 + 4n + 3} ) from above. Notice that for ( n \geq 1 ):

[ 2n^2 + 4n + 3 > 2n^2 ]

Hence:

[ \frac{5}{2n^2 + 4n + 3} < \frac{5}{2n^2} ]

Now, ( \sum_{n=1}^{\infty} \frac{5}{2n^2} ) is a p-series with ( p = 2 ), which is known to converge.

Therefore, by the Direct Comparison Test, since ( \frac{5}{2n^2 + 4n + 3} ) is bounded above by ( \frac{5}{2n^2} ) and ( \sum_{n=1}^{\infty} \frac{5}{2n^2} ) converges, ( \sum_{n=1}^{\infty} \frac{5}{2n^2 + 4n + 3} ) also converges.

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