# Does this serie converge ? $\sum_{n=1}^{\infty} \frac{1}{n^3 - 5n}$

Does this serie converge ? $$\sum_{n=1}^{\infty} \frac{1}{n^3 - 5n}$$

We have $$n^3 - 5n < n^3 \implies \frac{1}{n^3-5n} > \frac{1}{n^3}$$

We know $$\frac{1}{n^3}$$ converges because it's a p serie with $$p > 1$$, but how does that help for determining the convergence of $$\sum_{n=1}^{\infty} \frac{1}{n^3 - 5n}$$ ? What should I do next ?

• Asymptotically, you will have same behavior as $\frac{1}{n^{3}}$ . The precise statement is one known as the "Limit Comparison Test". Apply this and complete the solution. Oct 25, 2023 at 20:50
• As $~n~$ increases without bound, the denominator's $~5n~$ term is dwarfed by the denominator's $~n^3~$ term. So, informally (i.e. not valid without further analysis), this makes it game over. Oct 25, 2023 at 20:50
• @Mr.GandalfSauron Using the limit comparison test, we have $(1/n^3)/(1/(n^3-5n)) = (n^3 - 5n)/(n^3) = 1 - 5/n^2$ If we take $\lim_{n \to \infty} 1 - 5/n^2 = 1 = c < \infty$ so because we know the sum $1/n^3$ converges, our sum also has to converge, is that correct ? Oct 25, 2023 at 20:55
• To get a comparison going the correct direction, move a few terms out. So to get past $n^3-5n$ you might use $n\to n+2$ to get $n^3+6n^2+12n+1-5n=n^3+6n^2+7n+1$. Then peel off the number of terms to match, so you would manually evaluate $\frac 1{1^3-5\cdot 1}+\frac 1{2^3-5\cdot 2}$ and then compare the new sum as $\sum_{n=1}^{\infty}\frac 1{n^3+6n^2+7n+1}$ vs. the same using $\frac 1{n^3}$. Oct 25, 2023 at 20:56
• @wengen: yes, assuming I had calculated the transition correctly, you would be able to directly compare the two fractions and arrive at the conclusion (I should have had a $+8-10$ in the constant portion and so ideally I might have suggested using $n\to n+3$ instead, but here we are). Oct 25, 2023 at 21:58

For every $$n\ge3,$$ $$1-\frac5{n^2}\ge1-\frac5{3^2}=\frac49$$ hence $$\frac1{n^3-5n}\le\frac9{4n^3}.$$