I have two series, $S$ and $T$ which approximate $\pi$
such that $$S_n = 4 \sum_{i=1}^n \cfrac{-1^{i+1}}{2i-1}$$ and $$T_n = \Big(12 \sum_{i=1}^n \cfrac{-1^{1+i}}{k^2} \Big) ^{\frac{1}{2}}$$
It is known that $$|S_n - \pi| = \mathcal{O}(n^{-1})$$ and $$|T_n - \pi| = \mathcal{O}(n^{-2})$$
How can I show the truncation error, denoted by $E_t(n) = |S_n - \pi|$ is bounded such that $$\cfrac{2}{(n+2)^2} < E_t(n) < \cfrac{2}{n}$$
We have:
$$ \sum_{i=1}^{n}\frac{(-1)^{i+1}}{2i-1}=\sum_{j=0}^{n-1}\frac{(-1)^j}{2j+1}=\sum_{j=0}^{n-1}(-1)^j\int_{0}^{1}x^{2j}\,dx=\frac{\pi}{4}-(-1)^n\int_{0}^{1}\frac{x^{2n}}{1+x^2}\,dx $$ but over the interval $[0,1]$, the function $\frac{1}{1+x^2}$ is between $\frac{1}{2}$ and $1$, hence: $$\frac{1}{4n+2}\leq\left|-\frac{\pi}{4}+\sum_{i=1}^{n}\frac{(-1)^{i+1}}{2i-1}\right|\leq \frac{1}{2n+1}.$$