Proving the convergence of this series I have the series $\displaystyle\sum_{n=1}^{\infty}{\frac{6\sqrt{n}+5}{2n^2-n}}$ 
I am sure that this series converges, but I need to prove this and would like to use the comparison test to do so. 
I have rearranged 
$$
\frac{6\sqrt{n}+5}{2n^2-n} = \frac{6}{\sqrt{n}(2n-1)}+\frac{5}{2n-1} = \frac{1}{2n-1}(\frac{6}{\sqrt{n}}+5)
$$
but I am unsure how to use the comparison test in this case? 
 A: $6\sqrt{n} + 5 \leq 11 \sqrt{n}$ and $2n^2-n \geq n^2$. Hence,
$$\dfrac{6\sqrt{n} + 5}{2n^2-n} \leq \dfrac{11\sqrt{n}}{n^2} = \dfrac{11}{n^{3/2}}$$
Now conclude from the fact that $\sum \frac1{n^{3/2}}$ converges.
A: Using the asymptotic comparison
$$\frac{6\sqrt{n}+5}{2n^2-n}\sim_\infty \frac 3{n^{3/2}}$$
we see that the series is convergent since the Riemann series $\sum \frac 1{n^{3/2}}$ is convergent.
A: Instead of using the Basic Comparison Test, let us try and use another approach: The Limit Comparison test.
We are given the series $$\displaystyle \sum_{n=1}^\infty \frac{6\sqrt{n}+5}{2n^2 +n}$$
Let $\displaystyle a_n =\frac{6\sqrt{n}+5}{2n^2 +n} = \frac{1}{n^{\frac{3}{2}}}\times\frac{(6+\frac{5}{\sqrt{n}})}{(2+\frac{1}{n})} \ $ 
(Obtained by removing highest power in both numerator and denominator and simplifying)
Now choose $\displaystyle b_k = \frac{1}{n^{\frac{3}{2}}} \implies \sum_{k=1}^\infty b_k$ converges, since it is a Hyperharmonic Series with $p>1$.
Consider \begin{align}\frac{a_k}{b_k} &= \frac{6+\frac{5}{\sqrt n}}{2+ \frac{1}{n}} \xrightarrow{n \to \infty}\frac{6}{2}=3 >0\end{align}
Thus, by the Limit Comparison Test, we know $\displaystyle \sum_{n=1}^\infty \frac{6\sqrt{n}+5}{2n^2 +n}$ converges.
