Sequence Limits and the Squeeze Theorem

Evaluating limits of sequences that involve sums can be challenging, especially when the term inside the sum depends on both the index and the variable n. In this problem we'll use bounding inequalities to sandwich the given sequence between two simpler sequences, then apply the Squeeze Theorem to find the limit — a powerful technique frequently tested in HSC Extension 2.

Problem Statement

Let the sequence $u_n$ be defined by:

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

for integers $n \ge 1$. You may assume the standard result: $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$. (Do NOT prove this).

(i) Explain why, for all integers $k$ such that $1 \le k \le n$:

$$\frac{k^2}{n^3 + n} \le \frac{k^2}{n^3 + k} \le \frac{k^2}{n^3 + 1}$$

(ii) Hence, evaluate $\lim_{n \to \infty} u_n$.

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Hints

• Part (i): Consider how varying the denominator alters the overall value of a fraction. Since $k$ is positive, compare the sizes of $n^3+1$, $n^3+k$, and $n^3+n$.
• Part (ii): Apply the summation symbol $\sum_{k=1}^n$ across the inequality established in part (i). Factor out the denominators, substitute the given sum of squares formula, and apply the Squeeze Theorem to the outer bounds.

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Solutions

• Part (i): The key observation is that for $1 \le k \le n$, the denominator $n^3+k$ lies between $n^3+1$ and $n^3+n$, so the fraction $\frac{k^2}{n^3+k}$ will be sandwiched between the fractions with those outer denominators. For $1 \le k \le n$, the denominators satisfy $n^3 + 1 \le n^3 + k \le n^3 + n$. Taking the reciprocal reverses the inequality (as all terms are positive): $\frac{1}{n^3 + n} \le \frac{1}{n^3 + k} \le \frac{1}{n^3 + 1}$. Multiplying through by the positive term $k^2$ yields the required result: $\frac{k^2}{n^3 + n} \le \frac{k^2}{n^3 + k} \le \frac{k^2}{n^3 + 1}$.
• Part (ii): Now we sum the inequality from part (i) across all terms $k=1$ to $n$. Since $u_n$ is exactly the sum of the middle terms, this will bound $u_n$ between two sums that we can handle. Summing the inequalities from $k=1$ to $n$ gives: $\sum_{k=1}^n \frac{k^2}{n^3 + n} \le u_n \le \sum_{k=1}^n \frac{k^2}{n^3 + 1}$. Factoring out the denominators (since they are independent of $k$) yields $\frac{1}{n^3 + n} \sum_{k=1}^n k^2 \le u_n \le \frac{1}{n^3 + 1} \sum_{k=1}^n k^2$. Substituting the given sum formula gives $\frac{n(n+1)(2n+1)}{6(n^3+n)} \le u_n \le \frac{n(n+1)(2n+1)}{6(n^3+1)}$. To find the limit, we observe that as $n \to \infty$, the highest-degree term in both the numerator and denominator dominates, so we compare leading coefficients. As $n \to \infty$, the highest power of $n$ in both bounding expressions is $n^3$, leaving the ratio of the leading coefficients: $\frac{2}{6} = \frac{1}{3}$. Since both bounds approach $\frac{1}{3}$, by the Squeeze Theorem, $\lim_{n \to \infty} u_n = \frac{1}{3}$.

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Takeaways

• Bounding Strategies: Establishing inequalities by manipulating denominators is a fundamental technique for rigorous sequence problems, particularly well-suited for Extension 2 level assessments.
• Summation Mechanics: Recognizing that denominators independent of the summation index $k$ can be factored out of the sigma notation is a critical step that prevents algebraic clutter and confusion.
• Limit Theorems: This style of question seamlessly bridges the algebraic manipulation of series with formal limit evaluation using the Squeeze Theorem.