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.

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
$$ \eqalign{\sum_{n=N+1}^\infty \frac{1}{n^2} &< \frac{1}{N+1/2} - \int_{N+1}^\infty \dfrac{dx}{4x^4 - x^2}\cr
&= \frac{1}{N+1/2} - \ln\left(\frac{2N+3}{2N+1}\right)+ \frac{1}{N+1}}$$
