# Show that $\sum\limits_{n=1}^\infty\frac{2n^2-1}{3n^5+2n+1}$ converges or diverges

I'm working with some infinite series problems and I have to show that the series $\sum\limits_{n=1}^\infty\dfrac{2n^2-1}{3n^5+2n+1}$ converges or diverges. I don't have a lot of experience doing this yet, and this is a problem that my teacher made up so I have no way to check my answer.

For this problem, I said the series converges by direct comparison with $\dfrac{2}{3}\cdot\sum\limits_{n=1}^\infty\dfrac{1}{n^3}$ which converges by the p-series test. However, I'm not sure if I did this correctly and if all of the steps I took were "legal". This is my reasoning:

$\dfrac{2n^2-1}{3n^5+2n+1} \le \dfrac{2n^2}{3n^5}$ for all n.

$b_n = \dfrac{2n^2}{3n^5}=\dfrac{2}{3n^3}$

$\sum\limits_{n=1}^\infty\dfrac{2}{3n^3} =\dfrac{2}{3}\cdot\sum\limits_{n=1}^\infty\dfrac{1}{n^3}$ which converges by the p-series test.

Therefore, $\sum\limits_{n=1}^\infty\dfrac{2n^2-1}{3n^5+2n+1}$ converges by the direct comparison test with $\sum\limits_{n=1}^\infty\dfrac{1}{n^3}$.

Could anyone verify I used the test correctly or point out my mistakes? Thank you.

• Looks good to me. – Alex Becker Apr 4 '14 at 4:41
• $3n^5 \leq 3n^5 + 2n + 1$ and thus $\frac{2n^2 - 1}{3n^5 + 2n + 1} \leq \frac{2n^2 - 1}{3n^5} =\frac{2}{3}\frac{1}{n^3} - \frac{1}{3}\frac{1}{n^5}$ – Jared Apr 4 '14 at 4:45
• You're in luck with this rational function general term: the numerator is smaller than $\ 2n^2 \$ and the denominator is larger than $\ 3n^5 \$ for positive integers $\ n \$ , so you can use the direct (or simple) comparison test. Had the general term been $\ \frac{2n^2+1}{3n^5-2n+1} \ ,$ the inequality would run the wrong way. That's why we also have the Limit Comparison Test! – colormegone Apr 4 '14 at 4:55

You could just do limit comparison test with $\sum_{i=1}^{\infty} \frac{1}{n^{3}}$. I prefer LCT to DCT because you don't have to deal with inequalities anymore. XD
• Thanks for mentioning that - I actually tried the limit comparison test first, but thought it wasn't a valid test since I got 0. Isn't one of the conditions for the limit comparison test that $\lim\limits_{n\to\infty}\dfrac{a_n}{b_n}$ must be a finite number larger than 0? Perhaps I did the limit calculations wrong initially. – Sabien Apr 4 '14 at 5:29
• Wait, isn't it that $\lim_{n \to \infty} \frac{\frac{2n^2-1}{3n^5+2n+1}}{\frac{1}{n^3}} = \lim_{n \to \infty} \frac{2n^5-n^3}{3n^5+2n+1} = \frac{2}{3}$ ? – BCLC Apr 13 '14 at 11:57