The series $\sum_{n=1}^\infty \frac{n}{n+1}$ is convergent because it satisfies the limit comparison test. Comparing it to the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$, which is divergent, we have $\lim_{n\to\infty} \frac{\frac{n}{n+1}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{n^2}{n+1} = \infty > 1$. Therefore, $\sum_{n=1}^\infty \frac{n}{n+1}$ is also divergent.

The series $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$ is convergent because it is an alternating series that satisfies the alternating series test. The terms of the series decrease in absolute value, and the limit of the terms as $n\to\infty$ is 0. Therefore, the series converges.

The series $\sum_{n=1}^\infty \frac{1}{2^n}$ is a geometric series with first term $a = \frac{1}{2}$ and common ratio $r = \frac{1}{2}$. The sum of a geometric series is given by $S = \frac{a}{1-r}$. Substituting the values of $a$ and $r$, we get $S = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$. Therefore, the sum of the series is 1.

A series is absolutely convergent if the series of absolute values of its terms is convergent. The series $\sum_{n=1}^\infty \frac{1}{n^2}$ is absolutely convergent because the series $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent. The series $\sum_{n=1}^\infty \frac{n}{n+1}$ is not absolutely convergent because the series $\sum_{n=1}^\infty \frac{n}{n+1}$ is divergent.

The series $\sum_{n=1}^\infty \frac{n^2+1}{n^3+1}$ is convergent because it satisfies the limit comparison test. Comparing it to the series $\sum_{n=1}^\infty \frac{1}{n}$, which is convergent, we have $\lim_{n\to\infty} \frac{\frac{n^2+1}{n^3+1}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{n^3+n}{n^3+1} = 1$. Therefore, $\sum_{n=1}^\infty \frac{n^2+1}{n^3+1}$ is also convergent.

A telescoping series is a series in which most of the terms cancel out when they are added together. The series $\sum_{n=1}^\infty \frac{1}{n(n+1)}$ is a telescoping series because the terms can be rewritten as $\frac{1}{n} - \frac{1}{n+1}$, and when these terms are added together, most of them cancel out, leaving only the first and last terms.

The series $\sum_{n=1}^\infty \frac{1}{n(n+2)}$ is a telescoping series. Rewriting the terms as $\frac{1}{n} - \frac{1}{n+2}$, we have $\sum_{n=1}^\infty \frac{1}{n(n+2)} = \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+2}\right) = \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots = \frac{1}{2}$.

The series $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$ is divergent because it is an alternating series that does not satisfy the alternating series test. The terms of the series do not decrease in absolute value, and the limit of the terms as $n\to\infty$ is not 0. Therefore, the series diverges.

The series $\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$ is a telescoping series. Rewriting the terms as $\frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right)$, we have $\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left[\left(\frac{1}{1\cdot2} - \frac{1}{2\cdot3}\right) + \left(\frac{1}{2\cdot3} - \frac{1}{3\cdot4}\right) + \left(\frac{1}{3\cdot4} - \frac{1}{4\cdot5}\right) + \cdots\right] = \frac{1}{4}$.

A geometric series is a series in which the ratio of any two consecutive terms is constant. The series $\sum_{n=1}^\infty \frac{2^n}{3^n}$ is a geometric series because the ratio of any two consecutive terms is $\frac{2^{n+1}}{3^{n+1}} \div \frac{2^n}{3^n} = \frac{2}{3}$, which is constant.

The series $\sum_{n=1}^\infty \frac{2^n}{5^n}$ is a geometric series with first term $a = \frac{2}{5}$ and common ratio $r = \frac{2}{5}$. The sum of a geometric series is given by $S = \frac{a}{1-r}$. Substituting the values of $a$ and $r$, we get $S = \frac{\frac{2}{5}}{1-\frac{2}{5}} = \frac{2}{3}$.

The series $\sum_{n=1}^\infty \frac{n^2+2n+1}{n^3+3n^2+2n}$ is convergent because it satisfies the limit comparison test. Comparing it to the series $\sum_{n=1}^\infty \frac{1}{n}$, which is convergent, we have $\lim_{n\to\infty} \frac{\frac{n^2+2n+1}{n^3+3n^2+2n}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{n^3+3n^2+2n}{n^3+3n^2+2n} = 1$. Therefore, $\sum_{n=1}^\infty \frac{n^2+2n+1}{n^3+3n^2+2n}$ is also convergent.

The series $\sum_{n=1}^\infty \frac{1}{n(n+2)}$ is a telescoping series. Rewriting the terms as $\frac{1}{n(n+2)} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2}\right)$, we have $\sum_{n=1}^\infty \frac{1}{n(n+2)} = \frac{1}{2} \left[\left(\frac{1}{1\cdot3} - \frac{1}{3\cdot5}\right) + \left(\frac{1}{3\cdot5} - \frac{1}{5\cdot7}\right) + \left(\frac{1}{5\cdot7} - \frac{1}{7\cdot9}\right) + \cdots\right] = \frac{2}{3}$.

A harmonic series is a series of the form $\sum_{n=1}^\infty \frac{1}{n}$. The series $\sum_{n=1}^\infty \frac{1}{n}$ is a harmonic series.