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 $\frac{\pi^2}{6}$? As a math-for-fun exercise, I've recently been seeking bounds for the error $R_k$ made when using $\sum_{n=1}^{k}1/n^2$ to estimate its beautiful sum $\pi^2/6$. Applying the Comparison Test for series multiple times, I derived the following estimates:
$$\frac{1}{k+1}<\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k<\ln\left(1+\frac{1}{k}\right)<\frac{1}{2k+2}+\frac{1}{2k}$$
The lower estimate
$$\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k$$
which I derived with WolframAlpha's help, is pretty useless. $1/(k+1)<R_k$ is good, but I want to improve it. Given how clean and tight the upper estimate
$$R_k<\ln\left(1+\frac{1}{k}\right)$$
is, I figured I could find a similar lower estimate that's just as clean and tight. After thinking for a while, I came up empty handed. I can't seem to find a positive sequence $a_n$ lying between $1/(n^2+1)$ and $1/n^2$ for which $\sum_{n=1}^{k}a_n$ has a clean expression. Any ideas or hints?
Edit: I'm not trying to prove the convergence of $\sum_{n=1}^{\infty}\frac{1}{n^2}$ nor any other series.
 A: 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
$$ \eqalign{\sum_{n=N+1}^\infty \frac{1}{n^2} &< \frac{1}{N+1/2} - \int_{N+1}^\infty \dfrac{dx}{4x^4 - x^2}\cr
&= \frac{1}{N+1/2} - \ln\left(\frac{2N+3}{2N+1}\right)+ \frac{1}{N+1}}$$
A: 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 $$\Delta=\frac{207}{15190 N^8}+O\left(\frac{1}{N^9}\right)$$
A: It is known that for any $N,M\geq 1$
$$
\sum\limits_{n = N + 1}^\infty  {\frac{1}{{n^2 }}}  = \frac{1}{N} - \frac{1}{{2N^2 }} + \sum\limits_{m = 1}^{M - 1} {\frac{{B_{2m} }}{{N^{2m + 1} }}}  + \theta _M (N)\frac{{B_{2M} }}{{N^{2M + 1} }},
$$
where $B_m$ denotes the Bernoulli numbers and $0<\theta _M (N)<1$ is a suitable number depending on $N$ and $M$. From this you can obtain, for example,
$$
\sum\limits_{n = N + 1}^\infty  {\frac{1}{{n^2 }}}  > \frac{1}{N} - \frac{1}{{2N^2 }},\quad \sum\limits_{n = N + 1}^\infty  {\frac{1}{{n^2 }}}  > \frac{1}{N} - \frac{1}{{2N^2 }} + \frac{1}{{6N^3 }} - \frac{1}{{30N^5 }}.
$$
