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.

By Cauchy Schwarz we have $$\sum_{N\leq n\leq N+x}f(n)\leq\left(\sum_{N\leq n\leq N+x}\sqrt{f(n)}\right)^{2}\leq x\sum_{N\leq n\leq N+x}f(n).$$  Now, since these inequalities are best possible, that is since I can find $f$ with equality, or arbitrarily close to equality at either end, nothing more can be said without additional conditions on $f$.  
Notice that in particular, the left hand side gives $f$ diverges $\Rightarrow$ $\sqrt{f}$ diverges.
I mean, I flirted with the idea that $$\sum_{n=1}^\infty \frac{\sqrt{f}}{\sqrt{n}}$$ converging implies that $\sum_{n=1}^\infty f(n)$ must as well.  However, I think it is instructional to explain why this is not so:
Let $f(n)$ be the characteristic function of the fourth powers.
If monotonicity is also required, then this condition is true, but for general $f$, little can be said.
