I came across this infinite sum (or infinte product, or series?): $$\sum_{n=1}^{\infty }\frac{3n^2+7}{9n^3+4n-1}$$ which diverges to infinity. But how can I "prove" that? If the equation would be $$\sum_{n=1}^{\infty }\frac{9n^2+7}{3n^3+4n-1}$$ I could argue that the function is bigger than the Harmonic series $$\sum_{n=1}^{\infty }\frac{1}{n}$$ for any n and therefore has to diverge as well (Direct comparison test). But what could be a proper argument for the first equation? Is there a general rule for these "polynomial" equations?
PS: If my english is confusing please tell me, it's quite hard to translate these mathematical terms :)
The limit form of the comparison test: Suppose $a_n$ and $b_n$ are sequences with positive terms and suppose $\displaystyle \lim_{n\rightarrow \infty}{ a_n\over b_n}$ is finite and NOT zero, then $\displaystyle \sum^\infty a_n$ converges if and only if $\displaystyle\sum^{\infty} b_n$ converges.
So in this case,
As you noted you summands look like $1/n$ for large $n$, so choose $b_n={1\over n}$ and $a_n=\displaystyle { 9n^2+7\over 3n^3+4n^2-n}$.
Then $\displaystyle \lim_{n\rightarrow \infty}{ a_n\over b_n} =\displaystyle \lim_{n\rightarrow \infty}{ 9n^3+7n\over 3n^3+4n^2-n} =3 $ which is finite and non zero.
And since $\displaystyle \sum_{n=1}^{\infty}{ 1\over n} $ diverges and so does $\displaystyle\sum_{n=1} ^\infty a_n$.