To analyze the convergence of the given series, let's examine each series in detail using mathematical tests for convergence, focusing on whether they converge conditionally. Conditional convergence means the series converges, but it does not converge absolutely.

Analysis of Each Series:

1. Series 1:
   $\sum_{n=1}^\infty (-1)^n \ln^2\left(\cos\frac{1}{n}\right)$

   • As $n \to \infty$, $\cos\frac{1}{n} \to 1$ and $\ln(1) = 0$. Using a Taylor expansion, $\ln^2\left(\cos\frac{1}{n}\right) \sim \ln^2(1 - \frac{1}{2n^2})$, which decays slower than $\frac{1}{n^2}$.
   • Apply the Alternating Series Test: this series likely converges conditionally due to slow decay, but requires verification of absolute convergence.

2. Series 2:
   $\sum_{n=1}^\infty (-1)^n \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{3 \cdot 5 \cdot 7 \cdots (2n+1)}$

   • The terms decay as $\frac{1}{2n+1}$, which does not go to 0 fast enough for convergence.
   • Therefore, this series diverges.

3. Series 3:
   $\sum_{n=1}^\infty (-1)^n \frac{n^2 + (-1)^n}{3 + (-1)^n}$

   • The dominant term in the numerator and denominator is $n^2$, so the general term does not decay to 0.
   • This series diverges.

4. Series 4:
   $\sum_{n=1}^\infty (-1)^n n e^{-n}$

   • The term $n e^{-n}$ decays exponentially fast to 0.
   • This series converges absolutely, hence not conditionally.

5. Series 5:
   $\sum_{n=1}^\infty (-1)^n \frac{(n+2)}{n^2}$

   • Simplify: $\frac{n+2}{n^2} \sim \frac{1}{n}$ as $n \to \infty$.
   • This resembles the alternating harmonic series, which converges conditionally.

6. Series 6:
   $\sum_{n=1}^\infty (-1)^n \ln\left(\frac{n+1}{n}\right)$

   • Simplify: $\ln\left(\frac{n+1}{n}\right) = \ln(1 + \frac{1}{n}) \sim \frac{1}{n}$.
   • This resembles the alternating harmonic series, which converges conditionally.

7. Series 7:
   $\sum_{n=1}^\infty (-1)^n \frac{(2n)!}{1 \cdot 3 \cdot 5 \cdots (2n-1)}$

   • The general term grows factorially, so it diverges.

8. Series 8:
   $\sum_{n=1}^\infty (-1)^n \sin\frac{1}{n}$

   • As $n \to \infty$, $\sin\frac{1}{n} \sim \frac{1}{n}$.
   • This resembles the alternating harmonic series, which converges conditionally.

9. Series 9:
   $\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n(n+2)}}$

   • Simplify: $\frac{1}{\sqrt{n(n+2)}} \sim \frac{1}{n}$.
   • This resembles the alternating harmonic series, which converges conditionally.

10. Series 10:
    $\sum_{n=1}^\infty (-1)^n n^2 \tan\frac{1}{n^3}$

• For large $n$, $\tan\frac{1}{n^3} \sim \frac{1}{n^3}$, and the term becomes $n^2 \cdot \frac{1}{n^3} = \frac{1}{n}$.
• This resembles the alternating harmonic series, which converges conditionally.

Conclusion:

The series that converge conditionally are:

• Series 1
• Series 5
• Series 6
• Series 8
• Series 9
• Series 10

Let me know if you need a deeper explanation or want me to verify any calculations.

Related Questions:

1. How does the Alternating Series Test work for determining convergence?
2. What is the difference between absolute and conditional convergence?
3. How does exponential decay impact the convergence of series?
4. What role does factorial growth play in divergence of series?
5. How do Taylor expansions simplify convergence analysis for trigonometric functions?

Tip:

When analyzing alternating series, always check if the terms decay to 0 fast enough and compare their absolute convergence.