Let's break down and solve the two limit problems shown in the image:

(i) $\lim_{n \to \infty} \left[ \frac{1}{n} + \frac{1}{2n} + \frac{1}{3n} + \cdots + \frac{1}{n^2} \right] = 0$

This sum is of the form:

$S_n = \sum_{k=1}^{n} \frac{1}{kn}$

• The general term of this series is $\frac{1}{kn}$, where $k$ runs from 1 to $n$.
• This is equivalent to the sum:

$S_n = \frac{1}{n} \sum_{k=1}^{n} \frac{1}{k}$

The summation $\sum_{k=1}^{n} \frac{1}{k}$ is the harmonic series, which grows logarithmically as $n$ increases, i.e., $\sum_{k=1}^{n} \frac{1}{k} \approx \ln(n)$.

Thus, the expression becomes:

$S_n \approx \frac{1}{n} \ln(n)$

As $n \to \infty$, $\frac{\ln(n)}{n} \to 0$. Therefore, the limit is:

$\lim_{n \to \infty} S_n = 0$

(ii) $\lim_{n \to \infty} \left( 3^{-n} n^3 \right) = 0$

Here we need to evaluate:

$L = \lim_{n \to \infty} 3^{-n} n^3 = \lim_{n \to \infty} \frac{n^3}{3^n}$

• $n^3$ grows polynomially, while $3^n$ grows exponentially.
• Exponential growth always dominates polynomial growth as $n \to \infty$, meaning that $\frac{n^3}{3^n} \to 0$.

Thus, the limit is:

$\lim_{n \to \infty} \frac{n^3}{3^n} = 0$

Final Answers:

1. $\lim_{n \to \infty} \left[ \frac{1}{n} + \frac{1}{2n} + \frac{1}{3n} + \cdots + \frac{1}{n^2} \right] = 0$
2. $\lim_{n \to \infty} \left( 3^{-n} n^3 \right) = 0$

Would you like further details or clarifications on any part of this solution?

Here are 5 related questions to extend the learning:

1. What is the asymptotic behavior of the harmonic series?
2. How can we rigorously prove that exponential functions grow faster than polynomials?
3. Can you derive the approximation of the harmonic series $\sum_{k=1}^{n} \frac{1}{k} \approx \ln(n)$?
4. What are some other examples where polynomial growth is outpaced by exponential growth?
5. How do logarithmic terms influence the convergence of series?

Tip: When working with limits involving sums or series, it helps to approximate growth rates (polynomial, logarithmic, or exponential) to determine the behavior as $n \to \infty$.