= Solution
For $n\geq6$ and $n-1<x<n$, monotonicity gives $(x+1)^{-2}<n^{-2}<x^{-2}$. Integrating over each unit interval and summing proves the <integral test> tail bounds
$$
\boxed{\int_5^\infty\frac{dx}{(x+1)^2}=\frac16<\sum_{n=6}^\infty\frac1{n^2}<\int_5^\infty\frac{dx}{x^2}=\frac15.}
$$
The inequalities remain strict because the very first rectangle already has a strictly positive difference from either bounding <integral>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-1-reciprocal-square-rectangles.png]
{title=Unit rectangles between two shifted reciprocal-square curves}
{description=Original comparison for the first five intervals of the infinite tail. Each rectangle has height $1/n^2$ on $[n-1,n]$, above $(x+1)^{-2}$ and below $x^{-2}$.}
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