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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.

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    @check What does $P$ have to do with anything?2011-05-23

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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.

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    @List: My apologies!2011-05-23
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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 $a_k \geq 0$).

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    @Listing: Also, my reference to $a_n = O(n^{-2})$ was that it was *indeterminate*, i.e., (essentially) does not give anything useful.2011-05-23
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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.

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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 ${\rm E}(\sqrt{X}) \leq 1 + {\rm E}(X) < \infty$.)

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    @Eric: The probabilistic argument is perhaps more elementary here, since the inequlity ${\rm E}(\sqrt X ) \le \sqrt {{\rm E}(X)} $ is nothing but ${\rm Var}(\sqrt X ) \geq 0$.2011-05-24