convergence of series $\sum^{\infty}_{k=1}\frac{3k+1}{k^3+3k}$

Question

Finding whether the series is converge or diverge

$\displaystyle \sum^{\infty}_{k=1}\frac{3k+1}{k^3+3k}$

What i try:

$$\sum^{\infty}_{k=1}\frac{3k+1}{k^3+3k}<\sum^{\infty}_{k=1}\frac{3k+1}{k^3}=3\sum^{\infty}_{k=1}\frac{1}{k^2}+\sum^{\infty}_{k=1}\frac{1}{k^3}$$

Both series are converge using $p$ series test.

So original series is converge

Plesse explain me is my work is right. If not how do i solve it. Help me please

Answer 1

Your proof is fine. You can actually just note that $\frac{3k+1}{k^3+3k}\sim\frac{3}{k^2}$ for a comparison test.

Answer 2

For a crude approximation, but a working, one you may notice that for any $k\geq 1$ $$ 0<\frac{3k+1}{k^3+3k}<\frac{3}{k-\frac{2}{3}}-\frac{3}{(k+1)-\frac{2}{3}}$$ and $\frac{3}{k-2/3}\to 0$ as $k\to +\infty$, so $$ \sum_{k\geq 1}\frac{3k+1}{k^3+3k}\leq 1+\frac{3}{2-\frac{2}{3}} = \frac{13}{4}.$$