First of all, I am slightly confused by your notation. You seem to be mixing partial sums with series. Therefore, let's call $S$ and $R$ the following series:
$$
S = \sum_{i=1}^{\infty} a_i
$$
and
$$
R = \sum_{i=1}^{\infty} \sqrt{a_i},
$$
and let $S_n = \displaystyle\sum_{i=1}^{n} a_i$ and $R_n = \displaystyle\sum_{i=1}^{n} \sqrt{a_i}$ denote their $n^{th}$ partial sums. As has been pointed (most simply, by Listing) it is clear that if $S \to \infty$, then $R \to \infty$ as well. On the other hand, if the series $S$ converges fast enough that the ratio test applies:
$$
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1,
$$
then the series $R$ converges as well, again by the ratio test:
$$
\lim_{n \to \infty} \left|\frac{\sqrt{b_{n+1}}}{\sqrt{b_{n}}} \right| = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|^{1/2} < 1.
$$
This explains the examples $a_i = \frac{1}{2^i}$ and $a_i = \frac{1}{i^2}$. It is also good to keep in mind that if $a_i = \frac{1}{i^s}$, then $S$ converges whenever $s > 1$, and therefore $R$ converges whenever $s > 2$.
This certainly does not cover every case, but it is a good start.
It is convenient to represent every $q_n$ as a reduced fraction $\frac{a_n}{b_n}$. So $a_0=b_0=1$ and:
$$ \frac{a_{n+1}}{b_{n+1}} = \frac{\frac{a_n}{b_n}+2}{\frac{a_n}{b_n}+1}=\frac{a_n+2 b_n}{a_n+b_n}\tag{1}$$
or:
$$ \left(\begin{array}{c}a_{n+1}\\b_{n+1}\end{array}\right)=\left(\begin{array}{cc}1 & 2\\ 1 & 1\end{array}\right)\left(\begin{array}{c}a_{n}\\b_{n}\end{array}\right).\tag{2}$$
The eigenvalues of such a matrix are $1\pm\sqrt{2}$ and the eigenvectors are $ \left(\begin{array}{c}\pm\sqrt{2}\\1\end{array}\right)$.
Since $a_n,b_n>0$, $\lim_{n\to +\infty}\frac{a_n}{b_n}=\sqrt{2}$ follows.
Best Answer
Say the sequence $(a_n)$ defined by $a_{n+1} = f(a_n)$ for bijective $f$ converges to $a$.
If $f$ is continuous at $a$, then $$0 = \lim\limits_{n \to \infty} a_ {n+1}-a_n = \lim\limits_{n \to \infty} f(a_n)-a_n = f(a)-a$$
So $f(a)=a$. But then it also follows that $f^{-1}(a)=a$. So if the inverse sequence is started with seed value $b_0 = a$, the inverse sequence will converge also.
I'm not $100$% sure what can be said for such convergent recurrences where $f$ is not continuous at the limit.