convergence of a sum of two geometric 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 + .. = a1 + a2 + a3 + ...
where a2 is (1/2)^2 + (1/3)^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 after applications of the root and ratio test:
$lim_{n \rightarrow \infty} Inf\left(\frac{a_{n+1}}{a_n}\right) = 0 $
$lim_{n \rightarrow \infty} Sup\left(\frac{a_{n+1}}{a_n}\right) = +\infty $
$lim_{n \rightarrow \infty} Inf \sqrt[n]{a_n} = \frac{1}{\sqrt{3}} $
$lim_{n \rightarrow \infty} Sup \sqrt[n]{a_n} = \frac{1}{\sqrt{2}} $
I know that $\frac{(a_{n+1})}{(a_n)} = \frac{(3^{(n+1)} + 2^{(n+1)})}{(6.(3^n+2^n))}$  and $(a_n) = \frac{3^n + 2^n}{3^n . 2^n}$. But what to do afterwards ?
How to get these calculations ?
 A: So I believe you have decided to describe this series by, $$ \sum_{n = 1}^{\infty} \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n  $$
Root Test: 
Now by the expansion of the Binomial Theorem for each $n \in \Bbb N$, 
$$ 0 \lt \sqrt[n]{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n } \lt \sqrt[n]{\left({\frac{1}{2} + \frac{1}{3}}\right)^n } =  \left({\frac{1}{2} + \frac{1}{3}}\right)  =  \frac 5 6 $$
$$  0 \lt \lim \sqrt[n]{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n } \lt 1 $$
See this. From this it follows that our series converges. 
Ratio Test: 
First note that $  \left({\frac{1}{2}}\right)^{n + 1} + \left({\frac{1}{3}}\right)^{n + 1} =  \frac{1}{2}\left({\frac{1}{2}}\right)^{n} + \frac{1}{3}\left({\frac{1}{3}}\right)^{n} \lt \frac{1}{2}\left({\frac{1}{2}}\right)^{n} + \frac{1}{2}\left({\frac{1}{3}}\right)^{n} $ for each $n \in \Bbb N$
Therefore,
$$ 0 \lt \frac{ \left({\frac{1}{2}}\right)^{n + 1} + \left({\frac{1}{3}}\right)^{n + 1} }{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n} \lt \frac{1}{2}\frac{  \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n }{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n } = \frac{1}{2} $$
whence it follows that 
$$  0 \lt \lim \frac{ \left({\frac{1}{2}}\right)^{n + 1} + \left({\frac{1}{3}}\right)^{n + 1} }{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n} \lt \frac 1 2 \lt 1  $$
Clearly,  our series again converges. 

It is also probably worth mentioning that you can easily prove that if $\sum a_n$ and $\sum b_n$ are convergent series then $\sum a_n + b_n$ will converge too. And the sum will be the sum of the two convergent series. Therefore the convergence of the geometric series $\sum (\frac{1}{2})^n$ and $\sum (\frac{1}{3})^n$ imply the convergence of yours.  
