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
Best Answer
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.