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.
Best Answer
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.