8.(8pts) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. Circle the answer.
(1) $\sum _{n=1}^{\mathrm{\infty }}\ln (1+\frac{1}{n})$
(abs. converges, cond. converges, diverges.)
(2) $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{n+1}{n\sqrt{n}}$
(abs. converges, cond. converges, diverges.)
(3) $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\sin \left(\frac{1}{n^2}\right)$
(abs. converges, cond. converges, diverges.)
(4) $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{{2021}^n}{{2022}^n+{2020}^n}$
(abs. converges, cond. converges, diverges.)

Step 1 ：First, we will determine the convergence of each series using the appropriate tests:

Step 2 ：For series (1), we can use the comparison test.

Step 3 ：For series (2), we can use the limit comparison test with the series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{n}}$.

Step 4 ：For series (3), we can use the alternating series test.

Step 5 ：For series (4), we can use the ratio test.

Step 6 ：Applying these tests, we find the following results:

Step 7 ：The series $\sum _{n=1}^{\mathrm{\infty }}\ln (1+\frac{1}{n})$ diverges.

Step 8 ：The series $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{n+1}{n\sqrt{n}}$ is conditionally convergent.

Step 9 ：The series $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\sin \left(\frac{1}{n^2}\right)$ is absolutely convergent.

Step 10 ：The series $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{{2021}^n}{{2022}^n+{2020}^n}$ is absolutely convergent.

Step 11 ：$\boxed{{\displaystyle \text{Final Answer:}}}$

Step 12 ：$\boxed{{\displaystyle \text{(1) diverges}}}$

Step 13 ：$\boxed{{\displaystyle \text{(2) conditionally convergent}}}$

Step 14 ：$\boxed{{\displaystyle \text{(3) absolutely convergent}}}$

Step 15 ：$\boxed{{\displaystyle \text{(4) absolutely convergent}}}$