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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

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  • $\begingroup$ Write it as one power series and use the formula. $\endgroup$ Commented Mar 5, 2020 at 18:10
  • $\begingroup$ Stricto censu, $S$ is not a power series. $\endgroup$ Commented Mar 5, 2020 at 18:10
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    $\begingroup$ $S(x)\approx (1+x)\sum\frac{x^{4n+1}}{4n+1}$ so its radius of convergence = $R_1$ $\endgroup$ Commented Mar 5, 2020 at 18:12
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    $\begingroup$ Where does $R \ge \min(R_1,R_2)$ come from? Is it true? $\endgroup$
    – mjw
    Commented Mar 5, 2020 at 18:13
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    $\begingroup$ @mjw math.stackexchange.com/questions/3033970/… $\endgroup$
    – saulspatz
    Commented Mar 5, 2020 at 18:23

4 Answers 4

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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.

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

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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 $R=1$.

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

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