How to compute the $n^{th}$ partial sum of a series?

Question

Compute explicitly $S_n(x)$, the $n^{th}$ partial sum of the series

$$\sum_{k=1}^∞ \frac{x\left[-1+4k(k+1)x^2\right]}{(1+4k^2x^2)(1+4(k+1)^2x^2)}$$

then compute the sum $S(x)$ of the infnite series, and prove that, for $a > 0$, the series is not uniformly convergent on $(a, a)$, but is uniformly convergent on $(a, ∞)$

My attempt:

$$S_n(x) = \sum_{k=1}^∞ \frac?{1+4k^2x^2} - \frac?{1+4(k+1)^2x^2}$$ which is a telescoping series.

And then, having formed $S_n(x)$, I find $S(x)$ = $\lim_{n→∞} S_n(x)$.

Finally, I find $M_n = \sup|S_n(x) - S(x)|$ and if $\lim_{n→∞} M_n$ = $0$, then it converges uniformly. My problem is in the first step. I don't know how to compute $S_n(x)$ explicitly.help please? This is my attempt for partial fraction decomposition

Answer 1

Based on the discussion in the comments we are trying to do a partial fraction decomposition$$\frac{x\left[-1+4k(k+1)x^2\right]}{(1+4k^2x^2)(1+4(k+1)^2x^2)}=\frac{ax+b}{1+4k^2x^2} - \frac{cx+d}{1+4(k+1)^2x^2}\\=\frac{(ax+b)(1+4(k+1)^2x^2)-(cx+d)(1+4k^2x^2)}{(1+4k^2x^2)(1+4(k+1)^2x^2)}\\ -x+4k(k+1)x^3=(ax+b)(1+4(k+1)^2x^2)-(cx+d)(1+4k^2x^2)\\b+d=0\\4(k+1)^2b+4k^2d=0\\b=d=0\\a-c=-1\\4(k+1)^2a-4k^2c=4k(k+1)\\4(k+1)^2a-4k^2(a+1)=4k(k+1)\\(2k+1)a-4k^2=4k(k+1)\\a=\frac {8k^2+4k}{2k+1}\\a=4k\\c=4k+1$$