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.
 A: 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.
A: If $S_n$ is convergent you cannot say anything about $R_n$, for example if $a_n=1/n^2$ then $R_n$ diverges. If $a_n=1/2^n$ then $R_n$ converges too.
If $S_n$ diverges $R_n$ will diverge too because you have for $a < 1$ that $a < \sqrt{a}$ (This reasoning assumes that $a_k \geq 0$).
A: 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.
A: 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 ${\rm E}(\sqrt{X}) \leq 1 + {\rm E}(X) < \infty$.)
