Does $\sum_{n=0}^\infty \frac{3n}{n^2+2n}$ converge? My attempts:
Ratio-test: Inconclusive
Comparision-test (to harmonic/ geometric series):
$$\sum_{n=0}^\infty \frac{3n}{n^2+2n} = \sum_{n=0}^\infty \frac{3}{n+2} = 3 \cdot\sum_{n=0}^\infty \frac{1}{n+2}$$
The series is a little bit smaller than the harmonic series $$\sum_{n=0}^\infty \frac{1}{n}$$ and im therefore unsure.
 A: HINT: use that $$\frac{1}{n+2}\geq \frac{1}{2n}$$
A: Hint: Note that
$$
\frac{3n}{n^2 + 2n} \geq \frac{3n}{n^2 + 2n^2} = \frac 1n
$$
A: Compare it to $\sum_{n=2}^\infty\frac1n$ instead. Or note that the $3\cdot{}$ makes it larger than the harmonic series again.
A: The shortest way is, as often, with equivalents:  $\;n^2+2n\sim_\infty n^2$, so
$$\frac{3n}{n^2+2n}\sim_\infty\frac{3n}{n^2}=\frac3n,\quad\text{which diverges.}$$
A: Note: you can reindex:
$\sum_{n=0}^{\infty} \frac 1{n+2} = \sum_{m=2}^{\infty} \frac 1m = \sum_{m=1}\frac 1m -1$.
Which we can use to show $\sum_{n= 0}^{\infty} \frac 1{n+k}$ will always diverge no matter how large a $k$ is.  
$\sum\limits_{j=1}^{k-1}\frac 1j + \sum_\limits{n= 0}^{\infty} \frac 1{n+k}= \sum_\limits{j= 1}^{\infty}\frac 1j$.  We know $\sum\limits_{j=1}^{k-1}\frac 1j$ is a finite value, so $\sum_\limits{n= 0}^{\infty} \frac 1{n+k}$ converges if and only if $\sum_\limits{j= 1}^{\infty}\frac 1j$ does.  And we know it doesn't.
So you sum would only converge if the harmonic series coverged (and it would be $3(N - 1)$ if $N$ were the sum of the harmonic series).  But the harmonic series doesn't converge.
A: You can apply the Integral Test. Let $f(x)=\frac{3x}{x^2+x}$ be the function describing the series' n-th term. You can conclude that $f(x) > 0$ is continuous for every $x \in [1, +oo[$. You can derive the function and conclude that $\frac{d}{dx}f(x) < 0$ for the given interval. Notice that you have all the conditions satisfied to apply the test. Integrating $\int_{1}^{+oo}$ will lead you to conclude that the series diverges.
