Speeding up the convergence of a series

There are several methods to speed up the summation of series. For example Euler summation or the Shanks transformation. Here is a simple method that works quite well. Let $$f(n)=\frac{n^2+2n-1}{n^4+2n^2+1}$$ and $$g(n)=\tfrac{1}{2}f(2n-1)-f(2n)+\tfrac{1}{2}f(2n+1).$$ Then $$f(1)-f(2)+f(3)-\ldots = \tfrac{1}{2}f(1) +g(1)+g(2)+g(3)+\ldots$$ but the right hand side converges much faster than the left hand side. This is a generic method. For example if you take $$f(n)=\frac{1}{n}$$ then $$g(n)=\frac{1}{(2n-1)\cdot 2n \cdot (2n+1)}$$ and $$\log(2)=1-\frac{1}{2}+\frac{1}{3}-\ldots =\frac{1}{2}+\frac{1}{1\cdot 2\cdot 3} + \frac{1}{3\cdot 4 \cdot 5}+\ldots$$

Here is the first step toward making the series evaluate quicker. You may need just as many terms but it will be faster. Break the series up through partial fractions.

$$\dfrac {-x^{2}-2x+1} {x^{4}+2x^{2}+1} = -\frac{2 (x-1)}{\left(x^2+1\right)^2}-\frac{1}{x^2+1}$$

I will focus on the second term in the sum.

$$\sum_{x=0}^\infty \frac{(-1)^x}{x^2+1} = \sum_{n=0}^\infty \frac{1}{4n^2+1} - \sum_{k=0}^\infty \frac{1}{(2k+1)^2+1}$$

Each of the two sums can be evaluated by contour methods. This approach can be found in my post here. In short, for a sufficient rational function $f(z)$,

$$\lim_{N\to\infty} \sum_{k = -N}^{k = N} f(k)$$

is equal to the negative of the sum of the residues of $\pi f(z) \cot(\pi z)$ at the poles of $f(z)$. We can use this because we are summing an even function. This yields

$$\sum_{n=0}^\infty \frac{1}{4n^2+1} = \frac{1}{4} \left(2+\pi \coth \left(\frac{\pi }{2}\right)\right)$$

$$\sum_{k=0}^\infty \frac{1}{(2k+1)^2+1} = \frac{1}{4} \pi \tanh \left(\frac{\pi }{2}\right)$$

Subtract these (and simplify) to find

$$\sum_{x=1}^\infty \frac{(-1)^x}{x^2+1} = \frac{1}{2} (\pi \operatorname{csch}(\pi )-1)$$

Now we can conclude our first step by writing...

$$\sum _{x=1}^{\infty }\left( -1\right) ^{x}\dfrac {-x^{2}-2x+1} {x^{4}+2x^{2}+1} = -\frac{1}{2} (\pi \operatorname{csch}(\pi )-1) - \sum_{x=1}^\infty(-1)^x\frac{2 (x-1)}{\left(x^2+1\right)^2}$$

It may be tempting to try the same approach for the remaining series but we are not summing an even function so it will not be as easy.