# 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.

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tried to edit it to something meaningfull but I still don't get it. –  Arjang May 23 '11 at 14:06
@Arjang: I hope my edit helps. –  check123 May 23 '11 at 14:26
@check123: Do you actually assume that $a_k \geq 0$ (real)? –  Shai Covo May 23 '11 at 14:36
Listing's answer is the one I would have given. (Note that s/he assumes the terms are positive, which you have not explicitly done, but otherwise the roots will not be real numbers. Is this what you want?) For a series with positive terms and general term converging to zero, taking square roots will make the terms (eventually) a bit larger, in some cases by enough to convert a convergent series to a divergent series. I don't see a more general answer forthcoming than that... –  Pete L. Clark May 23 '11 at 14:41
@check What does $P$ have to do with anything? –  JavaMan May 23 '11 at 15:53
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First of all, I am slightly confused by your notation. You seems to be mixing partial sums with series. Therefore, let's call $S$ and $R$ the followings 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 rato 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 should be a good start.

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@DJK: My name is not Listed :-( –  Listing May 23 '11 at 16:26
@List: My apologies! –  JavaMan May 23 '11 at 16:29

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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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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Ah! I just added the $\frac{1}{n^2}$ example for further elaboration and here it features in the answer. I guess my edited question will help. –  check123 May 23 '11 at 14:28
@Listing: I believe you can say quite a bit more under your assumptions. What if $a_n = o(n^{-2})$, for example? $a_n = O(n^{-2})$ is, I believe, in fact, the indeterminate boundary case. –  cardinal May 23 '11 at 14:44
Yes I agree as we are talking about positive functions you can certainly find classes of $S_n$ that imply divergence or convergence of $R_n$ but that has no real practical use as you will (at least to my knowledge) not find a generic criterion for the convergence that doesn't go down to the specific case. –  Listing May 23 '11 at 16:22
@cardinal: $\sum \frac{1}{n^2 \log(n)}$ converges, $\sum \frac{1}{n \sqrt{\log(n)}}$ does not. –  nik May 23 '11 at 16:23
For your certain argument take $a_n=\frac{1}{n^2 \cdot ln(n)}$ then $a_n \in o(n^{-2})$ but it will still diverge. Also $a_n = O(n^{-2})$ doesn't give you anything useful. Edit: neat zulon I think we came up with the same idea in the same moment :-) –  Listing May 23 '11 at 16:24
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 ${\rm E}(\sqrt{X}) \leq 1 + {\rm E}(X) < \infty$.)
@Eric: The probabilistic argument is perhaps more elementary here, since the inequlity ${\rm E}(\sqrt X ) \le \sqrt {{\rm E}(X)}$ is nothing but ${\rm Var}(\sqrt X ) \geq 0$. –  Shai Covo May 24 '11 at 1:37