Limit Comparison Test on $\sum_{n=1}^{\infty}\frac{2n^2-1}{3n^5+2n+1}$ I have the series:
$\sum_{n=1}^{\infty}\frac{2n^2-1}{3n^5+2n+1}$
I compared it with:
$\sum_{n=1}^{\infty}\frac{n^2}{n^5}$
The limit is $\frac{2n^5-n^3}{3n^5+2n+1}$ as $n$ approaches infinity.
The limit works out to $\frac23$. I don't understand how this shows convergence? It is less than 1, so why is the p-series, the second series I have, converging?

You mixed up comparison test and limit comparison test.
Limit comparison test:
Consider two series of non negative reals $\sum  u_n$ and $\sum v_n$ if
$lim_{n\to \infty} \frac{u_n}{v_n}=l \space (0<l< \infty)$ then both the series converge or diverge simultaneously.

$u_n=\frac{2n^2-1}{3n^5+2n+1}$
Let $v_n=\frac{1}{n^3}$
Then $\lim_{n\to \infty}\frac{\frac{2n^2-1}{3n^5+2n+1}}{\frac{1}{n^3}}=\frac{2}{3}$
Since $\sum \frac{1}{n^3}$ converge ( $\sum\frac{1}{n^p}$ converges for $p>1$ ) , by limit comparison test $\sum\frac{2n^2-1}{3n^5+2n+1}$ also converge.
Note: Convergence of $p$- series.
