I was reading the history of the Basel problem in William Duhnam's book, Euler - The Master of Us All. The book tells how Jakob Bernouili did some clever manipulation to show that the sum of $1/n^2 < 2$, and that this was the best upper bound until Euler found the exact solution. By comparing the sequence to the integral of $1/x^2$, whose anti-derivative is $-1/x$, it is easy to get better estimates. The integral from $n$ to $n+1$ is between $1/(n+1)^2$ and $1/n^2$ and the integral from $n$ to $\infty$ is $1/n$. By taking the first $(n-1)$ or $n$ terms of the series and then using the integral to estimate the remainder, it is easy to show that $$ 1 + 1/2^2 + ... + 1/(n-1)^2 + 1/n < \text{series} < 1 + 1/2^2 + ... + 1/(n-1)^2 + 1/n^2 + 1/n. $$
The lower and upper bounds differ by $1/n^2$, so that by taking just $10$ terms you can get an estimate within .01 of the actual value. I am no Bernoulli but I easily bested his estimate. Why didn't Bernoulli do the same as I did?
The claim that $2$ was the best upper bound until Euler is almost surely an instance of a layman lack of understanding combined with exaggeration. I certainly expect people with rudimentary training in mathematics to easily obtain sharp estimates. In fact, Euler might have computed the limit to high accuracy to help him guess the closed form!
Just to prove that it's very trivial to obtain a good upper bound, note that $x \mapsto \frac{1}{x^2}$ is convex and hence $\sum_{k=a+1}^b \frac{1}{k^2} \le \int_{a+\frac{1}{2}}^{b+\frac{1}{2}} \frac{1}{x^2}\ dx = \frac{2}{2a+1}-\frac{2}{2b+1}$, and hence $\sum_{k=a+1}^\infty \frac{1}{k^2} \le \frac{2}{2a+1}$. For example, $\sum_{k=1}^\infty \frac{1}{k^2} \le \frac{1}{1}+\frac{1}{4}+\frac{2}{5} = 1.65$, which uses only $2$ terms but is already good to $2$ decimal places! Use $10$ terms and the error drops to about $0.00007$. To get a similarly good lower bound, simply use trapezoids instead of rectangles to bound a suitable integral. Again by convexity of $x \mapsto \frac{1}{x^2}$ we get $\frac{1}{2a^2}+\sum_{k=a+1}^{b-1} \frac{1}{k^2}+\frac{1}{2b^2} \ge \int_a^b \frac{1}{x^2}\ dx = \frac{1}{a}-\frac{1}{b}$, and hence $\sum_{k=a+1}^\infty \frac{1}{k^2} \ge \frac{2a-1}{2a^2}$. Using $3$ terms we get $\sum_{k=1}^\infty \frac{1}{k^2} \ge \frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{5}{18} \ge 1.638$. These bounds differ by $\frac{2}{2a+1}-\frac{2a-1}{2a^2} = \frac{1}{(2a+1)(2a^2)}$, and therefore we see immediately that using $n$ terms gives bounds to within $\frac{1}{4n^3}$.
Furthermore, the technique of telescoping series can easily be extended to great effect. Starting with $\frac{1}{n}-\frac{1}{n+1} = \frac{1}{n^2+n}$, the discrepancy is $\frac{1}{n^2}-\frac{1}{n^2+n} = \frac{1}{n^3+n^2}$ so we just need to add a suitable multiple of $\frac{1}{n^2}-\frac{1}{(n+1)^2} = \frac{2n+1}{n^2(n+1)^2}$ to cancel out the most significant part. Simply repeating this systematic algebraic procedure $c$ times gives an approximation that has error $O(\frac{1}{n^{2+c}})$ when using $n$ terms.
Also, it won't escape even an amateur that $\frac{1}{n-1}-\frac{1}{n+1} = \frac{2}{n^2-1}$, which is an approximation one order better than our initial estimate. To cancel the $O(\frac{1}{n^4})$ discrepancy efficiently, one can similarly use $\frac{1}{(n-1)^3}-\frac{1}{(n+1)^3} = \frac{6n^2+2}{(n^2-1)^3}$ which causes the discrepancy to drop to $O(\frac{1}{n^6})$. Concretely, we get $\sum_{n=a+1}^\infty \frac{1}{n^2} \in \sum_{n=a+1}^\infty \bigg( \frac{1}{2} \Big( \frac{1}{n-1}-\frac{1}{n+1} \Big) - \frac{1}{6} \Big( \frac{1}{(n-1)^3}-\frac{1}{(n+1)^3} \Big) + [-1,1](\frac{3}{n^6}) \bigg)$, which gives $\sum_{n=a+1}^\infty \frac{1}{n^2} \in \frac{1}{2} \Big( \frac{1}{a} + \frac{1}{a+1} \Big) - \frac{1}{6} \Big( \frac{1}{a^3} + \frac{1}{(a+1)^3} \Big) + [-1,1]\frac{3}{5a^5}$ by telescoping and the integral bound. With this, just $10$ terms would give $5$dp by theory alone, which is enough to start guessing the closed form. In fact one gets a value of about $1.6449325$.