I need someone to check my work please.
With regards to $A \in \mathbb{R}$, determine the radius of convergence and for $A = 0$, sum the order:
$$\sum_{n = 0}^{+\infty} = \frac{A(-1)^n + 2^{n+1}}{2^{n+1}(n+1)}\cdot x^{n+1}$$
Now, I have tried determining the radius of convergence:
$$R = \lim_{n \to +\infty} \left|\frac{a_n}{a_{n+1}}\right| = \lim_{n \to +\infty} \left|\frac{\frac{A(-1)^n}{2^{n+1}(n+1)}}{\frac{A(-1)^{n+1}}{2^{n+2}(n+2)}} + \frac{\frac{2^{n+1}}{2^{n+1}(n+1)}}{\frac{2^{n+2}}{2^{n+2}(n+2)}}\right| = \cdot\cdot\cdot=\lim_{n \to +\infty}\left|\frac{n+2}{(-1)(n+1)}\right| = 1 $$
Given that this is correct, I'm now supposed to check the convergence of the series for $x = \pm1$.
When I swap $x = 1$ in the series I get the sum of convergent ($\sum_{n = 0}^{+\infty}\frac{A(-1)^n}{2^{n+1}(n+1)}$, converges according to Leibniz test) and divergent $\left(\sum_{n=0}^{+\infty}\frac{2^{n+1}}{2^{n+1}(n+1)} \right)$ order, which is a divergent order.
For $x = -1$, I get $\sum_{n=0}^{+\infty}(-1)^{n+1}\frac{A(-1)^n + 2^{n+1}}{2^{n+1}(n+1)}$, which I think converges absolutely, according to D'Alembert test. So, my order converges absolutely for $x \in [-1,1)$ and diverges in all other cases.
For the summation: $$f(x) = \sum_{n = 0}^{+\infty}\frac{2^{n+1}}{2^{n+1}(n+1)}x^{n+1} = \sum_{n=0}^{+\infty}\frac{x^{n+1}}{n+1}$$ $$f'(x) = \left(\sum_{n=0}^{+\infty}\frac{x^{n+1}}{n+1}\right)' = \sum_{n=0}^{+\infty}\frac{(n+1)x^n}{n+1} = \sum_{n=0}^{+\infty}x^n = \frac{1}{1 - x}$$ $$\Downarrow$$ $$f(x) = \int_0^x\frac{\,dt}{1 - t} = -\ln|1 -x|$$
Thank you for reading this and for any suggestions you might have.
Some feedback First try to simplify the expression before applying the ratio test: notice that $(-1)^n=_\infty o(2^{n+1})$ then
$$\frac{A(-1)^n + 2^{n+1}}{2^{n+1}(n+1)}\cdot x^{n+1}\sim_\infty\frac{x^{n+1}}{n+1}$$ so clearly the radius is $1$. You thought that the series is absolutely convergent at $x=-1$? that's not true!