Question

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

Answer 1

By making the denominator smaller,

`5/{2n^2+4n+2} le 5/n^2`.

Since

`sum_{n=1}^infty 5/n^2=5sum_{n=1}^infty1/n^2`

is a convergent p-series with `p=2>1`,

`sum_{n=1}^infty {5}/{2n^2+4n+2}`

converges by Comparison Test.

I hope that this was helpful.