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

1 Answer

Oct 2, 2014

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.

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