On (absolute) convergence of $f_c:= c + \sum_{n=0}^{+ \infty} \frac{a_n}{n+1}x^{n+1}$ 
Let $R> 0$ and let $g: (-R,R) \longrightarrow \mathbb{R}$ be given by the convergent power series $$g(x):= \sum_{n=0}^{+ \infty} a_nx^n$$ for $|x| < R$. Let $c \in \mathbb{R}$ and let $f_c: (-R,R) \longrightarrow \mathbb{R}$ be given by $$f_c(x):=c+ \sum_{n=0}^{+ \infty} \frac{a_n}{n+1}x^{n+1}, \ |x| < R $$
  Problem: Show that $f_c$ is well defined

To show that $f_c$ is well defined I need to show that it is absolute convergent. At first glance it seemed like an easy exercise to me, but during the process I began to struggle. Notice that I did highlight that $g$ is defined as a convergent power series, it appears to be one of my main problems. 
My approach: Goal: Show that $f_c$ is absolutely convergent on $(-R,R)$ 
$$f_c(x) \leq \sum_{n=0}^N \frac{|a_n|}{n+1}|x|^{n+1} \leq \sum_{n=0}^N|a_n| |x|^n |x| \leq \sum_{n=0}^N |a_n| |x|^n R=R\sum_{n=0}^N |a_n||x|^n \leq R\sum_{n=0}^{+ \infty} |a_n| \cdot |x|^n$$
My idea was to show with the above inequalities that $f_c$ must be absolutely convergent by using that $g$ is convergent, however I fail to make the correct relationship between $f_c$ and $g$. It is given that $g$ is convergent but not absolutely convergent and I believe that the above statement would only hold true iff $g$ is also absolutely convergent. 
I was told that this approach is correct, however I don't feel right about it so far. I would appreciate some hints and insights. 
 A: Good question. Your worry about convergent vs. abosolutely convergent shows your attention to detail, which will serve you well in math.
First of all, on the left hand side, you should have $|f_c- c|$, not simply $f_c$.
There is a special fact about power series which guarantees that they will be absolutely convergent on the interior of their domain of convergence (which as you recall is an interval which may or may not contain the endpoints). It comes from considering the absolute value of the series a little differently, not as $\sum |a_nx^n|$, but as 
$$\sum (|a_n|^{1/n}|x|)^n$$
Now fix $x$. I claim that the above series converges if $$\limsup_{n\to \infty}|a_n|^{1/n}|x|<1,$$ 
diverges if $\limsup_{n\to \infty}|a_n|^{1/n}|x|>1$, and we don't know if $\limsup_{n\to \infty}|a_n|^{1/n}|x|=1$.
I can go into more detail here if you want. Anyway, this shows that if $|x|<\frac{1}{L}$, where $L = \limsup_{n\to \infty}|a_n|^{1/n}$, then the power series will converge absolutely, and $|x|>\frac{1}{L}$, it will not. So $\frac{1}{L}$ is your $R$!
