I was requested to determine if $$\sum_{n=1}^\infty\frac{1}{(2n+1)(2n-1)}$$ converges or diverges.
I know the following theorem $T_1$: if $f(x)$ is continuous, positive and decreasing in $[1, \infty)$ and $f(n)=a_n$, then
$a)$ if $\int_1^\infty f(x)$ converges, then $\sum_{n=1}^\infty a_n$ converges;
$b)$ if $\int_1^\infty f(x)$ diverges, then $\sum_{n=1}^\infty a_n$ diverges.
$Attempt:$
$i$) See that$$\frac{1}{(2n+1)(2n-1)}=\frac{A}{2n+1}+\frac{B}{2n-1} \implies 1=A(2n-1)+B(2n+1)$$
Solving for $A$ and $B$ we get $A=-\frac{1}{2}$ and $B=\frac{1}{2}$. Then
$$\frac{1}{(2n+1)(2n-1)}=\frac{-\frac{1}{2}}{2n+1}+\frac{\frac{1}{2}}{2n-1}$$
$ii)$ Let $f(x)=\frac{1}{(2x+1)(2x-1)}$ so that $f(n)=a_n=\frac{1}{(2n+1)(2n-1)}$. Notice that $f(x)$ is positive, increasing and continuous for all $x \geq 1$ (I'll skeep the proof of this for the sake of brevity).
See that $$\int_1 ^\infty \frac{1}{(2x+1)(2x-1)}dx=-\frac{1}{2}\int_1 ^\infty \frac{1}{2x+1}dx+\frac{1}{2}\int_1 ^\infty \frac{1}{2x-1}dx$$
$$=-\frac{1}{2}\lim_{t\to\infty}\int_1^t\frac{1}{2x+1}dx+\frac{1}{2}\lim_{t\to\infty}\int_1^t\frac{1}{2x-1}dx$$
$$=-\frac{1}{4}\lim_{t\to\infty}(\ln(2t+1)-\ln(3))+\frac{1}{4}\lim_{t\to\infty}(\ln(2t-1))$$
Since neither of the limits exist as finite numbers, $\int_1 ^\infty \frac{1}{(2x+1)(2x-1)}dx$ diverges.
$iii)$ Because of $T_1$, $\sum_{n=1}^\infty\frac{1}{(2n+1)(2n-1)}$ diverges too.
I have two main questions regarding this proof. First, is the procedure correct? Second, was there a simpler way to prove the divergence of the sequence without using improper integrals?
The manipulation $$\int_1 ^\infty \frac{1}{(2x+1)(2x-1)}dx=-\frac{1}{2}\int_1 ^\infty \frac{1}{2x+1}dx+\frac{1}{2}\int_1 ^\infty \frac{1}{2x-1}dx $$ is wrong because the $2$ integrals on the RHS are divergent and you have an indeterminate form $\infty -\infty$ on the RHS. Your mistake is similar to writing $\lim_{x\to\infty}(x-x)=\lim_{x\to\infty}x-\lim_{x\to\infty}x$ and concluding that $\lim_{x\to\infty}(x-x)$, which is $0$, doesn't exist since the $2$ limits on the RHS are infinite.
To show the convergence of the integral, you can use comparisons. For $x\geq 1$, we have $2x-1\geq x$ and $2x+1>x$, so $(2x-1)(2x+1) > x^2$. Thus $$\int_1^{\infty}\frac{1}{(2x-1)(2x+1)}\,dx<\int_1^{\infty}\frac{1}{x^2}\,dx<\infty $$ The convergence of the last integral is easier to show (if you don't know that already).