I would like to use the ratio and root test on the following series:

$$s = 1/2 + 1/3 + (1/2)^2 + (1/3)^2 + \ldots = a1 + a2 + a3 + \ldots$$

where $a_2$ is $\left(\frac{1}{2} \right)^2 + \left(\frac{1}{3} \right)^2 $ for example.

I know we have a sum of two geometric series so the sum will be convergent but I'd like to find the following results

$\lim_{n \to \infty} \inf\left(\frac{a_{n+1}}{a_n}\right) = 0 $

$\lim_{n \to \infty} \sup\left(\frac{a_{n+1}}{a_n}\right) = +\infty $

$\lim_{n \to \infty} \inf \sqrt[n]{a_n} = \frac{1}{\sqrt{3}} $

$\lim_{n \to \infty} \sup \sqrt[n]{a_n} = \frac{1}{\sqrt{2}} $

How to calculate such supremum of infimum ?

I know that $\frac{(a_{n+1})}{(a_n)} = \frac{(3^{(n+1)} + 2^{(n+1)})}{(6\cdot(3^n+2^n))}$ and $(a_n) = \frac{3^n + 2^n}{3^n \cdot 2^n}$. But what to do afterwards ?

How to get these calculations ?


Hint: Since $a_n=2^{-n}+3^{-n}$, we get $$ \begin{align} \frac{a_{n+1}}{a_n} &=\frac{2^{-n-1}+3^{-n-1}}{2^{-n}+3^{-n}}\\ &=\frac{1/2+1/3(2/3)^n}{1+(2/3)^n} \end{align} $$ and $$ \begin{align} a_n^{1/n} &=\left(2^{-n}+3^{-n}\right)^{1/n}\\ &=\frac12\left(1+(2/3)^n\right)^{1/n} \end{align} $$

To get the limits listed in the question, you may have meant to define $$ a_n=\left\{\begin{array}{} 2^{-k}&\text{if }n=2k-1\\ 3^{-k}&\text{if }n=2k \end{array}\right. $$ Then $$ \frac{a_{n+1}}{a_n}=\left\{\begin{array}{} (2/3)^k&\text{if }n=2k-1\\ 1/2(3/2)^k&\text{if }n=2k \end{array}\right. $$ and $$ a_n^{1/n}=\left\{\begin{array}{} 2^{-k/(2k-1)}&\text{if }n=2k-1\\ 3^{-k/(2k)}&\text{if }n=2k \end{array}\right. $$


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