Question

These convergence tests are slowly killing me. I need to find whether the following is absolutely convergent, conditionally convergent or divergent. And I can see it converges absolutely since it behaves like (2n^4)/(3n^6) but I can't figure out how to set up the comparison...and justify it

Answer 1

\[\sum_{n=1}^{\infty}(-1)^{n}\frac{ 2n^{4}-1 }{ 3n^{6}-n^{2}-1 }\]

Answer 2

So the aim is to set something up where that < a convergent p-series, but the -n^2 -1 in the denom. is confusing me

Answer 3

I realize that's what I'm supposed to do after comparing it to a p-series, it's just getting to it that's my problem D=

Answer 4

\(n^6 \gt n^2+1\) for all \(n\gt 1\)
\(\implies \dfrac{2n^4}{3n^6-(n^2+1)} \lt \dfrac{2n^4}{3n^6-(n^6)} \) for all \(n\gt 1\)

agree ?

Answer 5

Agreed and thank you!

Answer 6

for \(n\gt 1\) we have
\[\dfrac{2n^4-1}{3n^6-(n^2+1)} \lt \dfrac{2n^4}{3n^6-(n^2+1)} \lt \dfrac{2n^4}{3n^6-(n^6)} = \dfrac{1}{n^2}\]

next you know what to do