Question

Determine if the series converges or diverges.
\[\sum_{n=2}^{∞} \frac{4n^{2}-5}{3n^3+2n}\]

Answer 1

\[\sum_{n=2}^{∞} \frac{4n^{2}-5}{3n^3+2n}\]

Answer 2

Compare to the divergent series, \(\displaystyle \sum_{n=2}^\infty \frac{1}{n}\)

Answer 3

How do you decide that? is it because the highest order exponents simplify to 4/3n?

Answer 4

Yes

Answer 5

is this a scam

Answer 6

?

You just have to show that
\[\frac{4n^2-5}{3n^3+2n}>\frac{1}{n}\]

Answer 7

and if you divide out the ns you get 4/3n. How do I prove that \[\frac{4}{3n} > \frac{1}{n}\]

Answer 8

(4/3) times something is always bigger than that something.

Answer 9

oh wait I am severely mentally retarded forgive me

Answer 10

thank

Answer 11

Piss

Answer 12

yw