Comparison test for series $\sum_{n=1}^{\infty}\frac{n}{n^3 - 2n + 1}$ I am trying to prove the convergence of the series $\sum_{n=1}^{\infty}\frac{n}{n^3 -2n +1}$ with the simple comparison test. I know it can be done with other tests but this question came up in my homework for the comparison test before the other tests were discussed so I'm being stubborn and want to use that test.
I know it converges but I haven't been able to find a rigorous justification. This is my reasoning so far.
$\sum_{n=1}^{\infty}\frac{n}{n^3 -2n +1}$ converges because $\sum_{n=1}^{\infty}\frac{n}{n^3}$ = $\sum_{n=1}^{\infty}\frac{1}{n^2}$ and I know the latter converges by the p-series criterion.
To use the comparison test as my justification I must show $\frac{n}{n^3−2n+1} \leq \frac{n}{n^3} \ \ \forall \  n$. This would be obvious if the denominator on the left were greater than the denominator on the right but that can't be true as on the left I am subtracting from $n^3$ and on the right $n^3$ is unchanged.
So I switch to proving $\frac{n}{n^3−2n+1} \leq \frac{1}{n^2}$ and then I'm stuck.
If the numerators were both 1 then $n^3−2n+1$ is eventually going to be larger than $n^2$ even if we're subtracting a little bit from $n^3$.
That is, at n=1 we'll get $1^3−2+1=0 \leq 1$ but thereafter $n^3− 2n +1  \geq n^2$ is always true. 
But the statement of the comparison test says $a_n \leq b_n \forall n$ where $\sum b_n$ is known to converge.
So that's where I am so far. The problem comes down to showing the necessary inequality holds and is it enough the it holds "eventually".
Any ideas?  
 A: We need to start the summation somewhere after $n=1$, since the denominator is $0$ at $n=1$. We will show that $\sum_{n=2}^\infty \frac{n}{n^3-2n+1}$ converges. 
For large enough $n$, we have $n^3-2n+1\gt \frac{n^3}{2}$. This follows from general considerations. 
But more specifically, $n^3-2n+1\gt \frac{n^3}{2}$ for $n \ge 2$. This is because $n^3-2n\ge \frac{n^3}{2}$. To see that, note that the inequality $n^3-2n\ge \frac{n^3}{2}$ is equivalent to $\frac{n^3}{2}\ge 2n$, which is equivalent to $n^2\ge 4$. 
So for $n\ge 2$,  we have 
$$\frac{n}{n^3-2n+1} \lt \frac{2}{n^2}.$$
And we know that $\sum_2^\infty \frac{2}{n^2}$ converges. 
Remark: We used a common trick. If you have any polynomial $P(x)=a_0x^n +a_1x^{n-1}+\cdots$, where $a_0\ne 0$, then for large enough $|x|$, we have $|P(x)|\lt \left|\frac{a_0 x^n}{2}\right|$. The dominant term is $\dots$ dominant. 
A: How about the Limit Comparison Test? From Wikipedia:

Suppose that we have two series $\sum_n a_n$ and $\sum_n b_n$  with $a_n, b_n > 0$ for all $n$.
Then if $\lim_{n\to\infty}\frac{a_n}{b_n} = c$ with $0 < c < \infty$, then either both series converge or both series diverge.

For your problem, use:
$$a_n=\frac{n}{n^3 -2n +1}\\
b_n = \frac {1}{n^2}$$
Then:
$$\lim_{n\to\infty} \frac {a_n}{b_n} = \lim_{n\to \infty}\frac {\frac{n}{n^3 -2n +1}} {\frac 1 {n^2}} = \lim_{n\to \infty} {\frac{n^3}{n^3 -2n +1}} = 1\\$$
The last limit is a consequence of L'Hôpital's rule (see below).
Thus, because $0 < 1 < \infty$, and $\sum \frac 1 {n^2}$ converges (P-series), the given series must also converge.

The L'Hôpital's rule derivation is:
$\lim_{n\to\infty} \frac{n^3}{n^3 - 2n + 1} = \frac \infty \infty$ so we can take the derivative of the numerator and denominator.
$\lim_{n\to\infty} \frac{3n^2}{3n^2 - 2} = \frac \infty \infty$ so we can do it again.
$\lim_{n\to\infty} \frac{6n}{6n} = \frac \infty \infty$ so we do it once more.
$\lim_{n\to\infty} \frac{6}{6} = 1$ and we're done.
A: $\text{First, you should have your summation from $n=2$, since the denominator is $0$ for $n=1$.}$ $\text{Once you have this, note that}$
$$a_n = \dfrac{n}{n^3-2n+1} = \dfrac{n-1+1}{(n-1)(n^2+n-1)} = \dfrac1{n^2+n-1} + \dfrac1{(n-1)(n^2+n-1)}$$
$\text{Now note that for $n \geq 2$, we have}$
$n^2+n-1 > n^2$ and $(n-1)(n^2+n-1) > n^2$. $\text{Hence,}$
$$a_n < \dfrac2{n^2}$$
$\text{Now use the fact that }$$$\sum_{n=2}^{\infty} \dfrac1{n^p}$$ $\text{converges for $p>1$.}$
