Determine if this series $ \sum_{n=0}^\infty\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}$ converges Determine if the following series converges:
$$
\sum_{n=0}^\infty\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}.
$$
(http://i.stack.imgur.com/qWiuy.png)
I don't know how to  start.
 A: Use the limit comparison test. 
Let $\displaystyle a_n=\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}$ and $\displaystyle b_n=\frac{n^6}{n^7}=\frac{1}{n}$.
Since 
$$\begin{align}\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}&=\lim_{n\to\infty}\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}\cdot\frac{n}{1}\\
&=\lim_{n\to\infty}\frac{n^7\left(1+13\frac{1}{n}+\frac{1}{n^5}+\frac{1}{n^7}\right)}{n^7\left(1+13\frac{1}{n^3}+9\frac{1}{n^6}+\frac{2}{n^7}\right)} \\
&=1 > 0 \end{align}$$
Both $a_n$ and $b_n$ converges or diverges. Since $b_n$ diverges, so too must $a_n$.
A: Hint. You have, as $n \to \infty$,
$$
\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2} = \frac1n+\mathcal{O}{\left(\frac1{n^2} \right)}
$$ thus, by the comparison test, the initial series is divergent.
A: You want to use a comparison test. Remember with series that you can disregard a finite number of terms. When $n \geq 2$
$$n^7+13n^5+9n+2 \leq n^7 +13n^7 + 9n^7+n^2 = 24n^7,$$ 
which implies
$$ \frac{1}{n^7+13n^5+9n+2} \geq \frac{1}{n^7 +13n^7 + 9n^7+n^2} \geq \frac{1}{24n^7}.$$ 
For the numerator, you have 
$$n^6 + 13n^5 + n + 1 \geq n^6.$$
Therefore
$$ \sum_{n=2}^\infty \frac{n^6 + 13n^5 + n + 1}{n^6 + 13n^5 + 9n + 2} \geq \frac{1}{24} \sum_{n=2}^\infty \frac{n^6}{n^7} = \frac{1}{24}\sum_{n=2}^\infty \frac{1}{n} =\infty, $$
which shows that the series diverges by comparison with the harmonic series. 
