# 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 $$R=1$$ ??

• Write it as one power series and use the formula. Commented Mar 5, 2020 at 18:10
• Stricto censu, $S$ is not a power series. Commented Mar 5, 2020 at 18:10
• $S(x)\approx (1+x)\sum\frac{x^{4n+1}}{4n+1}$ so its radius of convergence = $R_1$ Commented Mar 5, 2020 at 18:12
• Where does $R \ge \min(R_1,R_2)$ come from? Is it true?
– mjw
Commented Mar 5, 2020 at 18:13
• Commented Mar 5, 2020 at 18:23

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 $$-1. By the way, if $$-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.

It looks like the radius of convergence is $$1$$, using Cauchy-Hadamard. We get $$r=1/\limsup_{n\to\infty}\sqrt[4n+1]{4n+1}=1$$.

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 $$R=1$$.

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 $$S(x)=\frac{1}{2} \left(\tan ^{-1}(x)+\tanh ^{-1}(x)+\tanh ^{-1}\left(x^2\right)\right)$$