Why is the radius convergence of $\sum_{n=0}^\infty \frac{x^{4n+1}}{4n+1}$ is $1$?

Question

Why is the radius convergence of $\sum_{n=0}^\infty \frac{x^{4n+1}}{4n+1}$ is $1$?

We know that

$$\frac{1}{R} = \limsup_{n\to\infty} \sqrt[n] {\frac{1}{4n+1}} = 0$$

And therefore, $R=\infty$. Why is it wrong?

Answer 1

$\displaystyle\dfrac{\frac{x^{4(n+1)+1}}{4(n+1)+1}}{\frac{x^{4n+1}}{4n+1}}\to_{n\to\infty}x^3$ hence the radius of convergence is $1$.

By the way, $\overline{\lim}_{n\to\infty} \sqrt[n] {\frac{1}{4n+1}} = \color{red}1$ :

$\left(\frac{1}n\right)^{1/n}=e^{\ln(1/n)/n}=e^{-\ln(n)/n}$ and $\ln(n)/n\to_{n\infty}0$

Answer 2

Note that the coefficient of the power series is $\frac1m$ whenever $m\equiv1\pmod4$, and $0$ otherwise, so using the Iverson bracket it can be written $\frac1m[m\equiv1\pmod4]$. Then what you should want to compute is $$ \limsup_{m\to\infty}\sqrt[m]{\frac{[m\equiv1\pmod4]}m}= \lim_{n\to\infty} \sqrt[n] {\frac1n} =\lim_{n\to\infty}\exp\left(\frac{-\ln n}n\right) =\exp(0)=1, $$ (since the growth of $n$ beats that of $\ln n$ as $n\to\infty$). This explains your error.