Direct comparison test for $\sum_{n=1}^{\infty}{\frac{n+4}{n^2-3n+1}}$

Question

I was wondering whether this is a valid argumentation:

Let $\sum_{n=1}^{\infty}{\underbrace{\frac{n+4}{n^2-3n+1}}_{:=a_n}}$ and $c_k:=\frac{n^2+4}{n^2-3n+1}$, then $$\frac{n+4}{n^2-3n+1}\leq \frac{n^2+4}{n^2-3n+1}=\frac{n^2\left(1+\frac{4}{n^2}\right)}{n^2\left(1+-\frac{3}{n}+\frac{1}{n^2}\right)}\longrightarrow 1$$ Since $a_k\leq c_k$ the series converges.

Answer 1

You have $c_k \rightarrow 1$, which means that $\sum_{n=1}^\infty c_n$ diverges. Therefore it's not enough to say that $\sum_{n=1}^\infty a_n$ convergent.

$\sum_{n=1}^\infty a_n$ is actualy divergent. Try showing that $a_n > \frac{1}{n}$ and that $\sum_{n=1}^\infty \frac{1}{n}$ is divergent.

Answer 2

Your argument makes no sense. With what convergent series are you comparing the given series?

$\frac {n+4} {n^{2}-3n+1} \geq \frac 1n$ and $\sum \frac 1 n$ is divergent so the series is divergent.