Bounding a sigma sum

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

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

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

Hints

[leftmargin=*]

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.

Solutions

(i) For $1 \le k \le n$ , the denominators satisfy

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

Taking reciprocals reverses the inequality (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 $k^2$ gives

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

(ii) Summing from $k=1$ to $n$ ,

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

Factoring out the $k$ -independent denominators,

\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 formula,

\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 numerator and denominator is $n^3$ , so both bounds tend to

\frac{2}{6}=\frac{1}{3}.

By the Squeeze Theorem,

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

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.

Further Readings

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