real analysis - Interchanging the integral and the infinite sum

I would like to know whether I can interchange the integral and the infinite sum as follows:

$$\int_{0}^{\infty}\sum_{k = 1}^{\infty} f_k(x)\mathrm{d}x = \sum_{k = 1}^{\infty}\int_{0}^{\infty} f_k(x)\mathrm{d}x,$$
where
$$f_k(x) = (-1)^{k+1}\binom{n}{k}\frac{c(ab)^{k}}{2(1+bx)^{k}}x^{k-c-1},$$
with $n \in \mathbb{C}$, $a \in (0,1]$, $b > 0$, and $c \in [0, 1]$.

Now according to Nate Eldredge's answer here, for general $f_k$ , if $\int \sum |f_k| < \infty$ or $\sum \int |f_k| < \infty$ (by Tonelli the two conditions are equivalent), then $\int \sum f_k = \sum \int f_k$.

My attempt:
I tried to prove the second condition as follows:
\begin{align}\sum_{k = 1}^{\infty}\int_{0}^{\infty} |f_k(x)|\mathrm{d}x &= \sum\limits_{k = 1}^{\infty} {\binom{n}{k}} \frac{c(ab)^{k}}{2} \int\limits_{0}^{\infty}\frac{x^{k-c-1}}{\left(1+b x \right)^{k}}\mathrm{d}x \\

&=\sum\limits_{k = 1}^{\infty} {\binom{n}{k}} \frac{c(ab)^{k}}{2} b^{c-k}\left[\frac{\pi}{\sin(\pi c)}\frac{\Gamma(k-c)}{\Gamma(k)\Gamma(1-c)}\right] \\
& = \frac{b^{c}}{2}\frac{\pi c}{\sin(\pi c)}\sum\limits_{k = 1}^{\infty} {\binom{n}{k}} a^{k} \left[\frac{\Gamma(k-c)}{\Gamma(k)\Gamma(1-c)}\right].\quad (1)
\end{align}

I could not prove that the last expression in $(1)$ is $< \infty$.

Is my attempt on the correct track? If yes, how to prove that the RHS of $(1)$ is $< \infty$? If no, is there an another way to prove that the integral and the infinite summation can be interchanged?

Answer

In order to prove that $(1)$ is finite you just have to compute the radius of convergence of a power series, namely
$$ an\,\phantom{}_2 F_1\left(1-c,1-n; 2;-a\right). $$

Since by the ratio test such radius of convergence equals one, the exchange of $\int$ and $\sum$ is allowed for any $a\in(0,1)$ and the $a=1$ case can be studied as a separate instance.

The exchange of $\int$ and $\sum$ in the $a\in(0,1)$ case then leads to the identity

$$ \int_{0}^{+\infty}cx^{-1-c}\left(1-\left(\frac{1+(1-a)bx}{1+bx}\right)^n\right)\,dx=cb^c\sum_{k\geq 1}(-1)^{k+1}a^k\binom{n}{k} B(c,k-c) $$
wher $B$ is Euler's Beta function. Both sides equals
$$ \frac{\pi nac b^c}{\sin(\pi c)}\phantom{}_2 F_1\left(1-c,1-n;2;a\right)=\frac{\pi n c b^c a(1-a)^{c+n}}{\sin(\pi c)}\phantom{}_2 F_1\left(c+1,n+1;2;a\right)$$
where the last identity is a consequence of Euler's transformations.