Given: $$ a_n= \begin{cases} \frac{1}{3^n} & \text{if $n$ is prime,}\\ \frac{1}{4^n} & \text{if $n$ is not prime}. \end{cases} $$ The ratio test will work fine here, but the way series is defined I am confused regarding it. Any help would be appreciated.

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    $\begingroup$ en.wikipedia.org/wiki/Root_test Note this is a limit superior ($\limsup$), not a simple limit. $\endgroup$ Nov 8, 2014 at 4:01
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    $\begingroup$ The $\limsup$ is equal to the supremum of the limits of the convergent subsequences (including tending to infinity as convergent). Then we can compute $\lim_{n\text{ is prime}}\sqrt[n]{|a_n|}=1/3$, and $\lim_{n\text{ is not prime}}\sqrt[n]{|a_n|}=1/4$. The supremum of $\{1/3,1/4\}$ is $1/3$. Therefore the radius of convergence is $3$. $\endgroup$
    – user152732
    Nov 8, 2014 at 13:43
  • $\begingroup$ @user152732 thanks a lot $\endgroup$ Nov 8, 2014 at 14:10
  • $\begingroup$ The comparison with the geometric series (which is the proof of the test) can illustrate how/why this is the case $\endgroup$
    – user152732
    Nov 8, 2014 at 14:52
  • $\begingroup$ @user152732 can you please explain ? i am new to this concept of power series $\endgroup$ Nov 8, 2014 at 14:53

1 Answer 1


Lets consider the power series $$p(x)=\sum_{n=1}^{\infty}a_nx^n,$$ where $$a_n= \begin{cases} \frac{1}{3^n} & \text{if $n$ is prime,}\\ \frac{1}{4^n} & \text{if $n$ is not prime}. \end{cases}$$ Then $p(x)$ lie between $$\min\{\sum_{n=1}^{\infty}\Big|\dfrac{x}{3}\Big|^n,\sum_{n=1}^{\infty}\Big|\dfrac{x}{4}\Big|^n\}$$ and $$\max\{\sum_{n=1}^{\infty}\Big|\dfrac{x}{3}\Big|^n,\sum_{n=1}^{\infty}\Big|\dfrac{x}{4}\Big|^n\}$$ with in the common radius of convergence of both $$\sum_{n=1}^{\infty}\Big|\dfrac{x}{3}\Big|^n,\,\,\,\,\,\sum_{n=1}^{\infty}\Big|\dfrac{x}{4}\Big|^n.$$ Also note that $$\sum_{n=1}^{\infty}\Big(\dfrac{x}{a}\Big)^n=\dfrac{a}{a-x}\iff|x|<|a|$$ and $\min\{3,4\}=3.$ Hence your power series has radius of convergence $r=3.$


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