The image contains three infinite series that we need to analyze for convergence or divergence. Let's examine each one step by step.

a) $\sum_{n=1}^\infty \frac{2n - 1}{\sqrt{2}^n}$:

• Rewrite the denominator: $\sqrt{2}^n = (2^{1/2})^n = 2^{n/2}$, making the series: $\sum_{n=1}^\infty \frac{2n - 1}{2^{n/2}}.$
• We need to determine convergence. For exponential terms like $2^{n/2}$, the ratio test or comparison test is useful.

b) $\sum_{n=1}^\infty \frac{3n}{(6n + 1)^n}$:

• Here, $(6n + 1)^n$ grows extremely fast compared to $3n$. This suggests rapid decay of the terms, so we suspect convergence.
• A common approach is to use the root test, as the terms are raised to the $n$-th power.

c) $\sum_{n=1}^\infty \frac{\ln^4(5n + 2)}{5n + 2}$:

• This series combines a logarithmic growth $\ln^4(5n + 2)$ and a linear denominator $5n + 2$.
• To assess convergence:
  • Compare it with $\sum \frac{\ln^4(n)}{n}$, which behaves like $\sum \frac{1}{n^{1 - \epsilon}}$ for large $n$.
  • For $p$-series with $p > 1$, convergence is likely.

Would you like detailed solutions for all three, or clarification on a specific one? Here are some related questions you might explore:

5 Related Questions:

1. How does the ratio test help determine convergence of series?
2. What are key differences between root and comparison tests?
3. How does logarithmic growth compare with polynomial or exponential growth?
4. What role do $p$-series play in convergence tests?
5. Why do terms involving $n!$ or $a^n$ often converge rapidly?

Tip:

For faster convergence tests, simplify the series terms to dominant growth forms (e.g., $n^k, \ln(n), a^n$), and compare with known series like geometric or $p$-series.