integral (2n^2 -1)/(3n^5 +2n+1) dn from 1 to infinity

Question

amistre64 · Answer

thats a rather odd looking rational expression

amistre64 · Answer

it mimics 2/3n^3, but that prolly doesnt help

amistre64 · Answer

as a dbl chk: http://www.wolframalpha.com/input/?i=integral+from+1+to+infinity+%282n%5E2+-1%29%2F%283n%5E5+%2B2n%2B1%29

amistre64 · Answer

by parts looks doable; if not lengthy ....

amistre64 · Answer

cant even see that

anonymous · Answer

i got 1/3 but the answer on wolfram is 0.206

amistre64 · Answer

what steps did you take?  numerical perhaps?

mr.math · Answer

I assume you're using this integral to test if the series $\sum_{n=1}^n \frac{2n^2 -1}{3n^5 +2n+1}$ converges, right?

mr.math · Answer

from n=1 to infinity*

amistre64 · Answer

if so then the comparison test might be good:
$$\lim_{n->inf}\  \frac{2n^2 -1}{3n^5 +2n+1}*\frac{3n^3}{2}=1$$
....well thats inconclusive

amistre64 · Answer

$$\frac{2n^2 -1}{3n^5 +2n+1}\ ?\le?\frac{2}{3n^3}$$
$$6n^5 -3n^2\ ?\le?6n^5 +4n+2$$
$$0\ \le\ 7n+2;\ n\ge0$$its true

so if 2/3n^3 converges; the lesser function has to as well

amistre64 · Answer

the 2/3 is rather moot; so 1/n^3 is a p-series, where the (p)ower is > 1 and so it converges