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.
