Convergence of infinite/finite 'root series' Let $S_n=a_1+a_2+a_3+...$  be a series where $ {a}_{k}\in \mathbb{R}$ and let $P = \{m\;|\;m\;is\;a\;property\;of\;S_n\}$ based on this information what can be said of the corresponding root series: $R_n=\sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ...$
In particular, if $S_n$ is convergent/divergent then in what circumstances can we say that $R_n$ is also convergent/divergent?
EDIT (1) 
Eg:
$$S_n = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$$ we know that the series converges to $1$. While the corresponding root series $$R_n = \frac{\sqrt{1}}{\sqrt{2}}+\frac{\sqrt{1}}{\sqrt{4}}+\frac{\sqrt{1}}{\sqrt{8}}+...$$ also converges (which we know does to $1+\sqrt2$).
We also know that the above convergence cannot generalised to all root series as, the series $\displaystyle \frac{1}{n^2}$ converges to $\displaystyle \frac{\pi^2}{6}$, while the corresponding root series $\displaystyle \sqrt{\frac{1}{n^2}}$ diverges.
My Question is: Is there a way to determine which 'root series' diverges or converges based only on information about the parent series.

It is interesting to make here the following observation.
While convergence of $\sum\nolimits_{k = 1}^\infty  {a_k }$ (where $a_k \geq 0$) does not imply convergence of $
\sum\nolimits_{k = 1}^\infty  {\sqrt {a_k } } $, it does imply convergence of $\sum\nolimits_{k = 1}^\infty  {\sqrt {p_k a_k } } $ for any sequence $(p_k)$ of positive numbers satisfying $\sum\nolimits_{k = 1}^\infty  {p_k }  = 1$, and it holds
$$
\sum\limits_{k = 1}^\infty  {\sqrt {p_k a_k } }  \le \sqrt {\sum\nolimits_{k = 1}^\infty  {a_k } } .
$$
This follows from the inequality
$$
{\rm E}(\sqrt{X}) \leq  \sqrt{{\rm E}(X)},
$$
where $X$ is a discrete random variable taking the value $a_k/p_k$ with probability $p_k$, and such that ${\rm E}(X) = \sum\nolimits_{k = 1}^\infty  {a_k }$ is finite. (The fact that finiteness of ${\rm E}(X)$ implies that of ${\rm E}(\sqrt{X}) = \sum\nolimits_{k = 1}^\infty  {\sqrt {p_k a_k } } $ follows simply from the trivial inequality $\sqrt{X} \leq 1 + X$, yielding ${\rm E}(\sqrt{X}) \leq 1 + {\rm E}(X) < \infty$.)
