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

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

• Do you know how to do partial fraction decomposition? – Clayton May 13 at 11:34
• @Clayton yes I do – pascale bou chahine May 13 at 11:43
• When you tried it here, did anything go wrong? – Clayton May 13 at 11:47
• @Clayton I'm gonna try it now and let you know – pascale bou chahine May 13 at 11:49
• The numerators in a partial fraction decomposition have to have a degree one less than the denominators, so the numerator of the first fraction should be $dx+e$ and the second $fx+g$. That gives you enough parameters. It looks to me like $e=g=0$ because the denominators have only even powers of $x$ and the numerators only odd. – Ross Millikan May 13 at 13:36

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