Speeding up the convergence of $\zeta(2)$ Let us denote by $S$ the sum of the series $\displaystyle\zeta(2)=1+\frac1{2^2}+\frac1{3^2}+\cdots$
Yes, I know (and you know) that $S=\frac{\pi^2}6$, but that is not relevant for the question that I am about to ask.
This series converges slowly. In fact, the sequence$$\left(S-\sum_{k=1}^n\frac1{k^2}\right)_{n\in\mathbb N}$$converges to $0$ at about the same rate as $\left(\frac1n\right)_{n\in\mathbb N}$. My question is about speeding up the rate of convergence of this series. More precisely, it is this: prove that there is a number $K\in(0,1]$ such that$$(\forall n\in\mathbb{N}):\left|S-\frac2{2n+1}-\sum_{k=1}^n\frac1{k^2}\right|\leqslant\frac K{n^3}.$$
Added note: Because of some comments that I got, I want to make this clear: I know an answer to this question.
 A: Obviously, the rest after the $n$th partial sum is $$S-\sum_{k=1}^n\frac1{k^2}=\sum_{k=n+1}^\infty\frac1{k^2}.$$ Let's approximate that with some similar series with known partial sums, so a telescoping series would be nice. A convenient choice would be $$\frac1{k^2-1/4}=\frac1{k-1/2}-\frac1{k+1/2},$$ so
$$\sum_{k=n+1}^\infty\frac1{k^2-1/4}=\frac1{n+1/2}=\frac2{2n+1}$$ is the main part. We're left with an estimate for the error, i.e. for $$\sum_{k=n+1}^\infty\left(\frac1{k^2-1/4}-\frac1{k^2}\right)=\sum_{k=n+1}^\infty\frac{1/4}{k^2(k^2-1/4)}.$$ Let's try with some telescoping series, too: We have $$\frac1{(k-1/2)^3}-\frac1{(k+1/2)^3}=\frac{3k^2+1/4}{(k^2-1/4)^3}\ge\frac{3}{(k^2-1/4)^2}\ge12\cdot\frac{1/4}{k^2(k^2-1/4)},$$ and this means $$\sum_{k=n+1}^\infty\frac{1/4}{k^2(k^2-1/4)}\le\frac{1/12}{(n+1/2)^3}.$$
A: Set
\begin{align}
a_m &= \sum_{m=1}^{m}\frac{1}{n^2}\\
b_m &= 2 \sum_{n=1}^{m}\frac{(-1)^{n+1}}{n^2}, \qquad m>1
\end{align}
Then perhaps prove or disprove
$$\bigg |\frac{\zeta(2)-b_n}{\zeta(2)-a_n} \bigg | \rightarrow 1$$
Whcih would imply $b_m$ converges a lot quicker than $a_m$.
A: From series
$$\pi^2 = 10 - \sum_{k=0}^\infty \frac{1}{((k+1)(k+2))^3}$$
we can write
$$\frac{\pi^2}{6}=\frac{5}{3}-\frac{1}{6}\sum_{k=0}^\infty \frac{1}{((k+1)(k+2))^3}$$
Equivalently,
$$\begin{align}
\frac{5}{3}-\frac{1}{6}\sum_{k=0}^{n} \frac{1}{((k+1)(k+2))^3} -\frac{\pi^2}{6}&= \frac{1}{6}\sum_{k=n+1}^\infty \frac{1}{((k+1)(k+2))^3}\\
&< \frac{1}{6}\sum_{k=n+1}^\infty \frac{1}{(k+1)^6}\\
&< \frac{1}{6}\frac{1}{5n^5}=\frac{1}{30n^5}\\
\end{align}$$
A: So, here is my own answer.
Since$$S-\sum_{k=1}^n\frac1{k^2}=\sum_{k=n+1}^\infty\frac1{k^2}\text{ and }\frac2{2n+1}=\sum_{k=n+1}^\infty\left(\frac2{2k-1}-\frac2{2k+1}\right)=\sum_{k=n+1}^\infty\frac4{4k^2-1}\text,$$we know that\begin{align*}\frac2{2n+1}-S+\sum_{k=1}^n\frac1{k^2}&=\sum_{k=n+1}^\infty\left(\frac4{4k^2-1}-\frac1{k^2}\right)\\&=\sum_{k=n+1}^\infty\left(\frac1{4k^4-k^2}\right).\end{align*}So\begin{align*}0&<-S+\frac2{2n+1}+\sum_{k=1}^n\frac1{k^2}\\&<\int_n^{+\infty}\frac{dx}{4x^4-x^2}\\&<\int_n^{+\infty}\frac{dx}{3x^4}\\&=\frac1{9n^3}.\end{align*}So, it starts like the answer provided by Professor Vector, but the method used to find an upper bound is different. I think that Professor Vector's answer is better than mine because it leads to a sharper estimate and because it doesn't require integrals.
