Decide if the series converges by using comparison test: $\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}+n}$ 
Decide if the series converges by using comparison test:
  $\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}+n}$

$$\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}+n}>\sum_{n=1}^{\infty}\frac{2n}{n^{2}}=2\sum_{n=1}^{\infty}\frac{1}{n}$$
Thus the series diverges because $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges.
Did I do it right?
 A: To lower bound a fraction, you can remove positive term from the numerator (which you did and is correct) but you cannot remove positive term from the denominator ! I would suggest the following bound :
$$
\frac{2n+1}{n^2 + n} > \frac{2n}{n^2 + n^2} = \frac{1}{n}$$
Then your conclusion about the harmonic series is correct.

EDIT : Some more information bounds for fraction :
To lower bound a fraction you either have to lower bound the top or upper bound the bottom, or both. In our case we have :
$$
2n + 1 > 2n \ \ \ \ \ \ \text{(lower bound of the top)} \\
n^2 + n < n^2 + n^2 \ \ \ \ \ \text{(upper bound of the bottom)}
$$
(For upper bounding it the opposite : upper bound the top or lower bound the bottom or both)
A: One other way to see it is to notice that
$${2n+1\over n^2+n}={n+1\over n^2+n}+{n\over n^2+n}={1\over n}+{1\over n+1}$$
And we're done
A: $\frac{2n+1}{n^2+n}>\frac{2n}{n^2}$ is equivalent with $2n^3+n^2>2n^3+2n^2$, which is equivalent with $n^2<0$, which doesn't hold. Your approach is wrong. 
