Does $\sum \limits _{k=1}^\infty \frac{3n+2}{n(n+1)}$ converges? Does $\sum \limits _{k=1}^\infty \frac{3n+2}{n(n+1)}$ converges?
Well, I know that $\frac{3n+2}{n(n+1)} = \frac{3}{(n+1)} + \frac{2}{n(n+1)}$
and I know that $\sum \limits _{n=1}^\infty \frac{3}{(n+1)}$ < 3 and $\sum \limits _{n=1}^\infty \frac{2}{n(n+1)}$ < 2 Which implies that (with arithmetic limits) that $\sum \limits _{k=1}^\infty \frac{3n+2}{n(n+1)}$ converges. Is that true? (In the book answers, it says that this series isn't converging, because it says that it converges to $\infty$ because $\sum \limits _{n=1}^\infty \frac{3}{(n+1)}$ converges to $\infty$ and the second one too, because it is always positive, which doesn't make sense with the solution I came up with.)
 A: You have
$$
\frac{3n+2}{n(n+1)}>\frac{3n}{n(n+1)}>\frac{1}{n+1} \quad \qquad n=1,2,\cdots,
$$ thus your initial series is divergent as is the harmonic series.
A: Your statement: $\sum_{n=1}^{\infty} \frac{3}{n+1} < 3$  is incorrect as the LHS goes to infitiny.
$$\sum_{n=1}^{\infty} \frac{3n+2}{n(n+1)} = \sum_{n=1}^{\infty} \bigg (\frac{1}{n+1} + \frac{2}{n} \bigg) >\sum_{n=1}^{\infty} \frac{2}{n}$$
which means your series is divergent.
A: It is obviously diverging, since the general term behaves like $\frac{3}{n}$.
$$\sum_{n=1}^{N}\frac{3n+2}{n(n+1)}=\sum_{n=1}^{N}\left(\frac{2}{n}+\frac{1}{n+1}\right)\geq 3H_N-1\geq 3\log N. $$
A: As it is a series with positive terms, equivalence* is enough:
$$\frac{3n+2}{n(n+1)}\sim_\infty\frac{3n}{n^2}=\frac3n,$$
which diverges
A: Say, $u_n=\frac{3n+2}{n(n+1)}$ and $v_n=\frac{1}{n}$
By comparison test, $$\lim_\limits{n\to\infty} \frac{u_n}{v_n}=\lim_\limits{n\to\infty} \frac{\frac{3n+2}{n(n+1)}}{\frac{1}{n}}=\lim_\limits{n\to\infty} \frac{3+\frac{2}{n}}{1+\frac{1}{n}}=3(\not =0,\infty)$$
And $\sum v_n$ is divergent.
So $\sum u_n$ is divergent.
