The problem is to evaluate the limit:

$\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + \ldots + n^2}{n^3}$

To solve this, let's break it down step-by-step.

Step 1: Sum of Squares Formula

The sum of the first $n$ squares is given by the formula:

$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

Step 2: Simplify the Expression

Substituting the sum of squares formula into the limit:

$\frac{1^2 + 2^2 + 3^2 + \ldots + n^2}{n^3} = \frac{\frac{n(n+1)(2n+1)}{6}}{n^3} = \frac{n(n+1)(2n+1)}{6n^3}$

Simplify this expression:

$= \frac{(n+1)(2n+1)}{6n^2}$

Further simplifying by expanding and dividing:

$= \frac{n^2 + 3n + 2}{6n^2} = \frac{1 + \frac{3}{n} + \frac{2}{n^2}}{6}$

Step 3: Taking the Limit as $n \to \infty$

As $n \to \infty$, the terms $\frac{3}{n}$ and $\frac{2}{n^2}$ approach zero:

$\lim_{n \to \infty} \frac{1 + \frac{3}{n} + \frac{2}{n^2}}{6} = \frac{1}{6}$

Thus, the value of the limit is:

$\boxed{\frac{1}{3}}$

This uses the interpretation of the sum as a Riemann sum. Specifically, the expression approximates the integral of $x^2$ over the interval $[0, 1]$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How can we generalize this limit problem to sums of higher powers?
2. What is the interpretation of this limit in terms of Riemann sums and integrals?
3. How do similar limits behave when the sum includes terms like $k^3$ instead of $k^2$?
4. How would this approach change if we were summing over different ranges or using different denominators?
5. Can we solve this problem using a different method, such as series expansion?

Tip: Understanding sums like these can help with integral approximations, especially when working with sequences and series that represent Riemann sums.