How to use p series to prove that this series converges So I know that the $p$ series converges if $p > 1$ in a series that looks like: 
$$\sum_{n=1}^\infty\frac{1}{n^p}$$   That's the p series.
So I have a question about this series: 
$$\sum_{n=4}^\infty \frac{n^{2.5} - 1}{n^{6.5} +4}$$
When I used cauchy condensation, I got that 
$$2^n \cdot \sum_{n=4}^{\infty}\frac{2^{2.5n}-1}{2^{6.5n} + 4}$$
I was wondering how I could reduce this to a p- series now and prove that it converges.
 A: by comparison test
$$\sum _{ n=4 }^{ \infty  } \frac { n^{ 2.5 }-1 }{ n^{ 6.5 }+4 } <\sum _{ n=4 }^{ \infty  } \frac { n^{ 2.5 }-1 }{ n^{ 6.5 } } <\sum _{ n=4 }^{ \infty  } \frac { 1 }{ n^{ 4 } } $$
A: More generally,
\begin{align}
\sum_{n=1}^\infty \frac{n^a +o(n^a)}{n^b +o(n^b)}
&=\sum_{n=1}^\infty \frac{n^a}{n^b}\frac{1+o(1)}{1+o(1)}\\
&=\sum_{n=1}^\infty  n^{a-b}(1+o(1))\\
\end{align}
converges if
$b > a+1$
and diverges if
$b \le a+1$.
This is the case
$a=2.5, b=6.5$,
so the sum converges.
A: Notice $$
\frac{n^{2.5} - 1}{n^{6.5} + 4}
\leq \frac{n^{2.5}}{n^{6.5} + 4} \leq \frac{n^{2.5}}{n^{6.5}} = \frac{1}{n^{4}}.
$$
The series $\displaystyle \sum_{n=4}^{\infty} \dfrac{1}{n^4}$ converges (because it is a $p$-series with $p = 4 > 1$). 
Therefore the series $\displaystyle \sum_{n=4}^{\infty} \dfrac{n^{2.5}-1}{n^{6.5}+4}$ converges by the comparison test.
Comparison Test: If $0 \leq a_n \leq b_n$ for all $n \geq N$ (for some fixed value $N$), and if $\sum_{n} b_n$ converges, then $\sum_{n} a_n$ converges.
In your example, $a_n = \dfrac{1}{n^4}$ and $b_n = \dfrac{n^{2.5}-1}{n^{6.5}+4}$.
