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?

  • 1
    $\begingroup$ How about the limit comparison test? $\endgroup$
    – wchargin
    Commented May 14, 2013 at 0:51
  • 1
    $\begingroup$ Are you sure the series starts at $n = 1$? The $n = 1$ term is undefined. Ignoring that, to see that the sum from $n = 2$ to $\infty$ is convergent, all you have to do is find a constant $c$ such that $\frac{n}{n^3 - 2n + 1} \leq \frac{c}{n}$. This gives you a bit more room for convergence testing. $\endgroup$
    – JavaMan
    Commented May 14, 2013 at 2:52

3 Answers 3


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.

  • $\begingroup$ I don't understand what an integrand has to do with the problem. And when you say "you want to start somewhere after n = 1,..." I don't understand what you mean. Start what? All I want to do is use the comparison test and that means I must verify the inequality $ a_n \ \leq \ b_n \forall \ n$ where $\sum b_n $ is known to converge. $\endgroup$
    – user77701
    Commented May 14, 2013 at 1:18
  • $\begingroup$ I should have said summand. By start I meant the sum can't start with $n=1$, since at $n=1$, the term $\frac{n}{n^3-2n+1}$ is not defined, since there $n^3-2n+1=0$. $\endgroup$ Commented May 14, 2013 at 1:28
  • $\begingroup$ @user77701: There has been some rewording, including removal of the wrong term integrand. $\endgroup$ Commented May 14, 2013 at 1:34
  • $\begingroup$ Ah, I understand. But.... the comparison test is always stated with the condition $ a_n \leq b_n \forall \ n$. The "for all" is what's tripping me up. It's very clear to me that $ a_n \leq b_n $ eventually. How do I demonstrate it with this specific test which has that strict condition "for all"? $\endgroup$
    – user77701
    Commented May 14, 2013 at 1:57
  • $\begingroup$ Actually, since our summation cannot start at $n=1$, and I proved the inequality holds for $n\ge 2$, we have shown it "for all." $\endgroup$ Commented May 14, 2013 at 2:02

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.

  • $\begingroup$ I appreciate you taking the time but my original question is how to do it specifically with the Comparison Test. I know it can be shown easily with other tests. I wish to understand how to do it with the Comparison Test. $\endgroup$
    – user77701
    Commented May 14, 2013 at 1:58
  • $\begingroup$ It turns out there was no way to do this with a direct comparison so your answer is the right one. Thanks. $\endgroup$
    – user77701
    Commented May 15, 2013 at 3:25

$\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 $p>1$.}$

  • $\begingroup$ Still lost. I understand why n = 1 can be ignored but don't understand at all what the purpose of that long equality at the top is and how it relates to the problem of showing $ a_n \leq b_n$ for all n that are defined. $\endgroup$
    – user77701
    Commented May 14, 2013 at 3:34

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