Let's analyze the three questions (8, 10, and 12) from the image and determine the convergence of the respective series.

Question 8

The series is: $\frac{1}{1 \cdot 2 \cdot 3} + \frac{3}{2 \cdot 3 \cdot 4} + \frac{5}{3 \cdot 4 \cdot 5} + \dots$

General term:

The general term can be written as: $a_n = \frac{2n - 1}{n(n+1)(n+2)}.$

Simplifying $a_n$:

Using partial fraction decomposition, we write: $\frac{1}{n(n+1)(n+2)} = \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)}.$ Thus, $a_n$ becomes: $a_n = (2n-1) \left( \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)} \right).$

Convergence Test:

For large $n$, the dominant term is $\frac{2n-1}{2n^3}$, which simplifies to approximately $\frac{1}{n^2}$. Since $\sum \frac{1}{n^2}$ converges (it is a p-series with $p > 1$), the given series converges.

Question 10

The series is: $\frac{2}{1^5} + \frac{3}{2^5} + \frac{4}{3^5} + \frac{5}{4^5} + \dots$

General term:

The general term is: $a_n = \frac{n+1}{n^5}.$

Simplifying $a_n$:

For large $n$, we approximate: $a_n \sim \frac{1}{n^4}.$

Convergence Test:

The term $\frac{1}{n^4}$ represents a p-series with $p = 4 > 1$. Hence, the series converges.

Question 12

The series is: $\sum_{n=1}^\infty \frac{\sqrt{n^2 - 1}}{n^3 + 1}.$

General term:

The general term is: $a_n = \frac{\sqrt{n^2 - 1}}{n^3 + 1}.$

Asymptotic behavior for large $n$:

For large $n$, $\sqrt{n^2 - 1} \sim n$, and $n^3 + 1 \sim n^3$. Thus: $a_n \sim \frac{n}{n^3} = \frac{1}{n^2}.$

Convergence Test:

Since the asymptotic term $\frac{1}{n^2}$ represents a p-series with $p = 2 > 1$, the series converges.

Summary

• Question 8: Converges
• Question 10: Converges
• Question 12: Converges

Do you want detailed derivations of any specific step?
Here are five related questions to deepen understanding:

1. How is partial fraction decomposition applied in series like Question 8?
2. What is the p-series test, and how is it applied to determine convergence?
3. Can alternating series converge even if their terms do not decrease fast enough?
4. What are the conditions for comparing two series using the comparison test?
5. How can asymptotic approximations simplify the analysis of series for convergence?

Tip: Always simplify the general term and compare it to a standard convergent or divergent series for quicker results!