For definiteness, call the terms of our sequence $a_1,a_2,a_3,\dots$. A similar analysis with minor differences of detail can be made if we call the first term of our sequence $a_0$.
Note that for $n=1,2,3,\dots$ we have $a_{2n-1}=\dfrac{1}{2^n}$ and $a_{2n}=\dfrac{1}{3^n}$.
The $k$-th root of the $k$-th term is "small" when the $k$-th term is a power of $\dfrac{1}{3}$. The $k$-th root of the $k$-th term is "large" when the $k$-th term is a power of $\dfrac{1}{3}$.
More precisely, $\liminf \sqrt[k]{a_k}=\lim\inf \sqrt[2n]{\frac{1}{3^n}}=\dfrac{1}{3}$. For even $k$ the $k$-th root is constant.
Also, $\limsup\sqrt[k]{a_k}=\liminf\sqrt[2n-1]{\dfrac{1}{2^n}}$. But
$$\sqrt[2n-1]{\dfrac{1}{2^n}}=\left(\frac{1}{2^n}\right)^{1/(2n-1)}=\left(\frac{1}{2^n}\right)^{2n/(2n(2n-1))}=\left(\frac{1}{\sqrt{2}}\right)^{2n/(2n-1)}.$$
The expression on the right has limit $\dfrac{1}{\sqrt{2}}$.
That takes care of one of the gaps.
For the Ratio Test, we are interested in the behaviour of $\left|\dfrac{a_{k+1}}{a_k}\right|$.
Let $k$ be odd, say $k=2n-1$. Then $a_k=\dfrac{1}{2^n}$. And $a_{k+1}=a_{2n}=\dfrac{1}{3^n}$. It follows that
$$\frac{a_{k+1}}{a_k}=\frac{a_{2n}}{a_{2n-1}}=\left(\frac{2}{3}\right)^n.$$
This has very pleasant behaviour for large $n$, indeed for any $n$: it is safely under $1$, indeed has limit $0$.
Now let $k$ be even, say $k=2n$. Then $a_k=\dfrac{1}{2^n}$. and $k+1=2n+1$. The $2n+1$-th term of our sequence is $\dfrac{1}{2^{n+1}}$. It follows that in the case $k=2n$ we have
$$\frac{a_{k+1}}{a_k}=\frac{a_{2n+1}}{a_{2n}}=\frac{\frac{1}{2^{n+1}}}{\frac{1}{3^n}}=\frac{1}{2}\left(\frac{3}{2}\right)^n.$$
This unfortunately behaves badly for large $n$: we would like it to be safely under $1$, and it is very much over.
The limit of the ratios $\dfrac{a_{k+1}}{a_k}$ does not exist. The ratios do not (uniformly) blow up, since for $k$ odd, the ratios approach $0$. The ratio behaves very nicely at odd $k$, and very badly at even $k$. So the Ratio Test is inconclusive. The bad behaviour prevents us from concluding convergence. But the good behaviour prevents us from concluding divergence.
It can be shown that
$$\liminf \frac{a_{n+1}}{a_n}\le \liminf \sqrt[n]{a_n}\le \limsup \sqrt[n]{a_n}\le \limsup\frac{a_{n+1}}{a_n}$$
therefore convergence by ratio test implies convergence by root test but not viceversa.
Refer also to the related:
For the series we can use that
$$\sum_{n=3}^\infty \left(\frac{1}{2}\right)^{\lceil\log_2 n\rceil-1}
=\sum_{n=1}^\infty \,\sum_{k=2^{n}+1}^{2^{n+1}} \left(\frac{1}{2}\right)^{n}
=\sum_{n=3}^\infty(2^{n}-1)\left(\frac{1}{2}\right)^{n}$$
Best Answer
I suppose that there is some confusion here. Indeed, the series$$\sum_{k=1}^\infty\left(\frac1{2^k}+\frac1{3^k}\right)$$converges and its convergence can be proved through the ratio test. My guess, is that the series whose convergence you ought to be studying is$$\frac12+\frac13+\frac1{2^2}+\frac1{3^2}+\frac1{2^3}+\frac1{3^3}+\cdots,\tag1$$which is the same as the one that you have mentioned, but without grouping terms in pairs. And, as far as the series $(1)$ is concerned, it is indeed true that it converges, but the ratio test is inconclusive: the superior limit is $\infty$, whereas the inferior limit is $0$.