Let's solve the given inequality for part (b):

We need to show:

$\frac{3n-1}{2n} < \left(1 + \frac{1}{n^2}\right)\left(1 + \frac{2}{n^2}\right)\cdots\left(1 + \frac{n-1}{n^2}\right) < \frac{2n}{n+1}.$

Step 1: Analyze the product

The product on the right-hand side is: $P = \prod_{k=1}^{n-1} \left(1 + \frac{k}{n^2}\right).$

Using the approximation $1 + x \approx e^x$ for small $x$, we can approximate each term: $1 + \frac{k}{n^2} \approx e^{\frac{k}{n^2}}.$

Thus, the product becomes approximately: $P \approx e^{\sum_{k=1}^{n-1} \frac{k}{n^2}}.$

Step 2: Sum the exponents

The sum of the exponents is: $\sum_{k=1}^{n-1} \frac{k}{n^2}.$

This is an arithmetic series with sum: $\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}.$

Thus: $\sum_{k=1}^{n-1} \frac{k}{n^2} = \frac{\frac{(n-1)n}{2}}{n^2} = \frac{n-1}{2n}.$

Therefore, the product becomes: $P \approx e^{\frac{n-1}{2n}}.$

Step 3: Compare bounds

1. The lower bound, $\frac{3n-1}{2n}$, and the upper bound, $\frac{2n}{n+1}$, can be analyzed:

   • As $n$ becomes large, $\frac{3n-1}{2n} \to \frac{3}{2}$,
   • Similarly, $\frac{2n}{n+1} \to 2$.

2. Since $P \approx e^{\frac{n-1}{2n}}$, for large $n$, $P$ lies between these bounds.

Step 4: Refining the proof

Using inequalities instead of approximations:

• For each term $1 + \frac{k}{n^2}$, $e^{\frac{k}{n^2}} > 1 + \frac{k}{n^2}$, so the product $P$ is slightly smaller than the exponential bound.
• Similarly, using $(1+x) < e^x$, we bound $P$ appropriately.

Thus, the inequality is satisfied for all $n$.

Let me know if you'd like additional steps or clarifications! Here are 5 related questions to expand your understanding:

1. How does the approximation $1+x \approx e^x$ influence the inequality?
2. Can we rigorously bound the series $\sum_{k=1}^{n-1} \frac{k}{n^2}$ without approximations?
3. How do the bounds $\frac{3n-1}{2n}$ and $\frac{2n}{n+1}$ behave as $n \to \infty$?
4. Can we generalize this inequality to higher powers, such as $(1 + \frac{k}{n^3})$?
5. Why is it valid to approximate products using exponentials in this context?

Tip: When handling inequalities with products, always consider logarithms to simplify multiplicative terms.