Q:

Convergence of $\sum \frac{k+1}{3k^3+2k^2+k}$

Does the series $\sum \frac{k+1}{3k^3+2k^2+k}$ converge?
Basically this series behaves like $\frac{1}{3}\sum \frac{k}{k^3} = \frac{1}{3}\sum \frac{1}{k^2}$ which we know converges (this can be proved by using the integral test). To make this formal we use the comparison test.
For any $k\geq 1$ we have
\begin{equation*}
k< k+1,
\end{equation*}
and so
\begin{equation*}
\frac{k+1}{3k^3+2k^2+k}> \frac{k}{3k^3+2k^2+k}> \frac{k}{3k^3}= \frac{1}{3k^2}
\end{equation*}
for all $k\geq 1$. But the series
\begin{equation*}
\frac{1}{3}\sum \frac{1}{k^2}
\end{equation*}
converges, so by the comparison test $\sum \frac{k+1}{3k^3+2k^2+k}$ converges.
This doesn't seem to be correct (in particular the inequality). Do I have to factorise the bottom of the fraction and compare with a different function? Any help would be good!!!

A:

You bounded your series from below by a convergent one, that isn't helpful. The second inequality is also false.
Comparison is the way to go however. Note that $k+1\leq 2k$ for $k\geq 1$, so that 
$$
\frac{k+1}{3k^3+2k^2+k}\leq \frac{2k}{3k^3+2k^2+k}\leq \frac{2k}{3k^3+2k^2}=\frac{2}{3k^2+2k}\\
\leq\frac{2}{3k^2}
$$
which is summable.

