Show that \(\sum\limits_{k=1}^n \frac{1}{k^2}\) is bounded above by a constant.

Intuitively this summation is bound above by the constant \(2\).

Using the techniques from this appendix section, note that we are dealing with a monotonically decreasing function and so we can use a form of (A.12) to show this.

\[\begin{split} \sum\limits^n_{k=1} \frac{1}{k^2} &= 1 + \sum\limits_{k=2}^n \frac{1}{k^2} \\ &\leq 1+ \int_1^n \frac{dx}{x^2} \\ &= 1 + \left( - \left.\frac{1}{n}\right|^{n}_1 \right)\\ &= 1 + \left(- \frac{1}{n} - \left( - \frac{1}{1}\right) \right) \\ &= 1 - \frac{1}{n} + 1 \\ &= 2 - \frac{1}{n} \\ &\leq 2 \\ \end{split}\]