Radius of convergence of a power series!! We consider the following Power series:
$$S(x)=\sum_{n\geq0} \frac{x^{4n+1}}{4n+1}+ \sum_{n\geq0} \frac{x^{4n+2}}{4n+2}.$$
I try to calculate the radius of convergence $R$ of $S(x)$. 
I know that the convergence radius of a sum of two power series of radius $R_1$ and $R_2$ is $\geq \min(R_1, R_2)$. Using Alembert's formulae, we obtain $R_1=R_2=1$, then $R\geq \min(R_1, R_2)=1$. But I don't know is it $R=1$ ??
Thank you in advance
 A: First off all $S(x)$ is defined as sum of two power series so you can treat it as one series if and only if they are both convergent. A rigorous proof that the set of convergence is $(-1,1)$ could be the following.
You have that with $-1< x<1$ both the series are convergent and therefore it is also their sum. With $x>1$ they are bot positive divergent and therefore it is also their sum. With $x=-1$ the first series is convergent by Leibniz Test while the second series is positive divergent so it is also their sum. With $x<-1$ let us consider
$$
\begin{gathered}
  S_n (x) = \sum\limits_{k = 0}^n {\frac{{x^{4k + 1} }}
{{4k + 1}} + } \sum\limits_{k = 0}^n {\frac{{x^{4k + 2} }}
{{4k + 2}} = }  \hfill \\
   \hfill \\
   = \sum\limits_{k = 0}^n {\frac{1}
{x}\frac{{x^{4k + 2} }}
{{4k + 1}} + } \sum\limits_{k = 0}^n {\frac{{x^{4k + 2} }}
{{4k + 2}} = }  \hfill \\
   \hfill \\
   = \sum\limits_{k = 0}^n {x^{4k + 2} \left( {\frac{1}
{x}\frac{1}
{{4k + 1}} + \frac{1}
{{4k + 2}}} \right)}  \hfill \\ 
\end{gathered} 
$$
If
$$
a_k (x) = x^{4k + 2} \left( {\frac{1}
{x}\frac{1}
{{4k + 1}} + \frac{1}
{{4k + 2}}} \right)
$$
we have that
$$
\mathop {\lim }\limits_{k \to  + \infty } \left| {a_k (x)} \right| =  + \infty 
$$
thus$S_n(x)$ does not converges as $n \to +\infty$. This proves that the $S(x)$ takes real values if and only if $-1<x<1$. By the way, if $-1<x<1$ we have that 
$$
S(x) = \sum\limits_{k = 0}^{ + \infty } {x^{4k + 1} \left( {\frac{1}
{{4k + 1}} + \frac{x}
{{4k + 2}}} \right)} 
$$
which is not a power series. This shows that not only power series admit a convergence's radius.
A: It looks like the radius of convergence is $1$, using Cauchy-Hadamard.  We get $r=1/\limsup_{n\to\infty}\sqrt[4n+1]{4n+1}=1$.
A: As you say, the radius of convergence is at least $1$.  Both series diverge at $x=1$, and since in that case they are series of positive of terms, so also does their sum.  Therefore, the radius of convergence cannot be $>1$, and we have $R=1$.
A: This is not an answer.
If you know the Taylor series, you should be able to identify that
$$\sum_{n=0}^\infty \frac{x^{4n+1}}{4n+1}=\frac{1}{2} \left(\tan ^{-1}(x)+\tanh ^{-1}(x)\right)$$ which already shows the result (divergence for $x=1$).
For the second summation
$$\sum_{n=0}^\infty \frac{x^{4n+2}}{4n+2}=\frac 12\sum_{n=0}^\infty \frac{(x^2)^{2n+1}}{2n+1}=\frac 12\tanh ^{-1}\left(x^2\right)$$
All of the above make
$$S(x)=\frac{1}{2} \left(\tan ^{-1}(x)+\tanh ^{-1}(x)+\tanh ^{-1}\left(x^2\right)\right)$$ 
