<s>[INST] Inf and Sup for ratio and root test of convergent series 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 ?
 [/INST] 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
$$
a_n^{1/n}=\left\{\begin{array}{}
2^{-k/(2k-1)}&\text{if }n=2k-1\\
3^{-k/(2k)}&\text{if }n=2k
\end{array}\right.
$$
</s>