4

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.

Daniel W. Farlow
  • 22,531
Sophie Clad
  • 2,291
  • 4
    http://en.wikipedia.org/wiki/Root_test Note this is a limit superior ($\limsup$), not a simple limit. – Andrés E. Caicedo Nov 08 '14 at 04:01
  • 1
    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 $3$. –  Nov 08 '14 at 13:43
  • @user152732 thanks a lot – Sophie Clad Nov 08 '14 at 14:10
  • The comparison with the geometric series (which is the proof of the test) can illustrate how/why this is the case –  Nov 08 '14 at 14:52
  • @user152732 can you please explain ? i am new to this concept of power series – Sophie Clad Nov 08 '14 at 14:53
  • For example, the series of the absolute values of the given series is $\sum |a_n||x|^n$. This is $\leq \sum |x|^n/3^n$. Therefore, whenever the latter converges the first is going to converge too. If the first converges, then the original series converges absolutely. For the geometric series $\sum |x/3|^n$ we know that it converges for $|x/3|<1$, i.e. |x|<3. Since the original series diverges for $x=3$, in fact $\sum a_n3^n$ has infinitely many terms equal to $1$, we know the radius can't be larger than $3$. –  Nov 08 '14 at 14:58

1 Answers1

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 $r=3.$

Bumblebee
  • 18,220
  • 5
  • 47
  • 87