Bounding a sigma sum

In HSC Advanced Mathematics, bounding techniques are essential for evaluating limits of sequences defined by sums. This problem shows how clever manipulation of denominator inequalities can turn a complicated sum into a simple limit, using the Squeeze Theorem and the familiar formula for the sum of squares. You’ll practise turning a sum with variable denominators into two sums with constant denominators, then letting (n \to \infty).

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

$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}. \text{(Do NOT prove this.)}$$

• Explain why, for all integers $k$ such that $1 \le k \le n$,

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

• Hence, evaluate $\lim_{n \to \infty} u_n$.

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Hints

• Part (i): Since $k$ is positive, compare the sizes of $n^3+1$, $n^3+k$, and $n^3+n$.
• Part (ii): Sum the inequality from part (i) over $k=1$ to $n$. Factor out denominators independent of $k$, substitute the sum-of-squares formula, and apply the Squeeze Theorem.

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Solutions

(i) We first compare the denominators for a fixed $n$ and an index $k$ between $1$ and $n$. For $1 \le k \le n$, the denominators satisfy

$$n^3 + 1 \le n^3 + k \le n^3 + n.$$

Because all terms are positive, taking reciprocals reverses the inequality:

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

Multiplying through by the positive term $k^2$ preserves the direction, giving the required bounds:

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

(ii) Now we sum the inequality over all $k$ from $1$ to $n$ to bound $u_n$:

$$\sum_{k=1}^n \frac{k^2}{n^3 + n} \le u_n \le \sum_{k=1}^n \frac{k^2}{n^3 + 1}.$$

The denominators $n^3+n$ and $n^3+1$ do not contain the summation index $k$, so we can factor them out:

$$\frac{1}{n^3 + n} \sum_{k=1}^n k^2 \le u_n \le \frac{1}{n^3 + 1} \sum_{k=1}^n k^2.$$

We substitute the standard sum-of-squares formula to express everything in terms of $n$ alone:

$$\frac{n(n+1)(2n+1)}{6(n^3+n)} \le u_n \le \frac{n(n+1)(2n+1)}{6(n^3+1)}.$$

As $n \to \infty$, the highest power in both numerator and denominator is $n^3$, so the rational bounds behave like $\frac{2n^3}{6n^3} = \frac{1}{3}$ for large $n$. More precisely, both bounds converge to

$$\lim_{n\to\infty} \frac{n(n+1)(2n+1)}{6(n^3+n)} = \frac{2}{6} = \frac{1}{3}, \qquad \lim_{n\to\infty} \frac{n(n+1)(2n+1)}{6(n^3+1)} = \frac{1}{3}.$$

With both the lower and upper bound approaching $\frac{1}{3}$, the Squeeze Theorem forces $u_n$ to have the same limit:

$$\lim_{n \to \infty} u_n = \frac{1}{3}.$$

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Takeaways

• Bounding by manipulating denominators is a core technique for rigorous sequence limits.
• Denominators independent of the summation index $k$ can be factored out of sigma notation.
• This style of question links algebraic series manipulation with limit evaluation via the Squeeze Theorem.

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Further Readings

HSC Collections, HSC Vectors, HSC Combinatorics, HSC Distributions