Is there a simple, but tight lower bound for the error made when $\sum_{n=1}^{k}\frac{1}{n^2}$ is used to approximate $\frac{\pi^2}{6}$?

Consider $$g(n) = \frac{1}{n-1/2} - \frac{1}{n+1/2}$$ Then $$ g(n) - \frac{1}{n^2} = \frac{1}{4n^4 - n^2} > 0 \ \text{for}\ n \ge 1$$

Now $1/(4n^4 - n^2)$ is a decreasing function of $n$ for $n > 1$, so $$\eqalign{\sum_{n=N+1}^\infty \frac{1}{n^2} &= \frac{1}{N+1/2} - \sum_{n=N+1}^\infty \frac{1}{4n^4-n^2}\cr & > \frac{1}{N+1/2} - \int_{N}^\infty \frac{dx}{4x^4 - x^2} \cr &= \frac{1}{N+1/2} - \ln \left(\frac{2N+1}{2N-1}\right) + \frac{1}{N}}$$ while on the other side $$ \eqalign{\sum_{n=N+1}^\infty \frac{1}{n^2} &< \frac{1}{N+1/2} - \int_{N+1}^\infty \dfrac{dx}{4x^4 - x^2}\cr &= \frac{1}{N+1/2} - \ln\left(\frac{2N+3}{2N+1}\right)+ \frac{1}{N+1}}$$

In the same spirit as @Robert Israel in his answer, we could use $$\sum_{n=N+1}^\infty \frac{1}{n^2}=\psi ^{(1)}(N+1)$$ and use the series expansion of the rhs $$\psi ^{(1)}(N+1)=\frac{1}{N}-\frac{1}{2 N^2}+\frac{1}{6 N^3}-\frac{1}{30 N^5}+\frac{1}{42 N^7}-\frac{1}{30 N^9}+O\left(\frac{1}{N^{11}}\right)$$ which, since alternating, allows to propose as sharp bounds as required.

What could be interesting is to look at the expansion of @Robert Israel's results $$\frac{1}{N+\frac 12} - \log \left(\frac{2N+1}{2N-1}\right) + \frac{1}{N}=\frac{1}{N}-\frac{1}{2 N^2}+\frac{1}{6 N^3}-\frac{1}{8 N^4}+O\left(\frac{1}{N^5}\right)$$ $$\frac{1}{N+\frac 12} - \log \left(\frac{2N+3}{2N+1}\right) + \frac{1}{N+1}=\frac{1}{N}-\frac{1}{2 N^2}+\frac{1}{6 N^3}+\frac{1}{8 N^4}+O\left(\frac{1}{N^5}\right)$$

Much less accurate : in my former group, we used for numerical purposes the simple double inequality $$\sinh \left(\frac{1}{N+1}\right)<\psi ^{(1)}(N+1)<\frac{1}{2} \sinh \left(\frac{2}{N}\right)$$

Later, playing with Padé-like approximants, I found better bounds (much better for the lower than for the upper) $$\color{blue}{\frac{3 (2 N+1)}{2 \left(3 N^2+3 N+1\right)}<\psi ^{(1)}(N+1)}<\frac{2 N^2+7 N+7}{2 (N+1)^2 (N+2)}$$ $$\Delta=\psi ^{(1)}(N+1)-\frac{3 (2 N+1)}{2 \left(3 N^2+3 N+1\right)}=\frac{1}{45 N^5}+O\left(\frac{1}{N^6}\right)$$

Continuing for answering this question, a better one $$\color{blue}{\frac{5N(1302 N^2+573 N+697) } {6(1085 N^4+1020 N^3+910 N^2+285 N+27 ) }<\psi ^{(1)}(N+1)}$$ and for this one $$\Delta=\frac{207}{15190 N^8}+O\left(\frac{1}{N^9}\right)$$