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I am trying to understand the concept that a sum of a positive series converges iff the sum of a function of the series converges, i.e.

$\sum a_n$ converges $\iff$ $\sum f(a_n)$ converges for $a_n >0$.

My book has questions applying this theorem, but does not explicitly state it in the chapter, or provide a proof. Would someone be able to provide a sketch of the proof, or explain why we have the requirement $a_n >0$?

I would like to show the above expressions specifically for the case that $f(x)=sin(x)$. Note that this does hold for $f(x)=sin(x)$ and $a_n$ as described, but does not hold for all functions.

EDIT: My professor just told me that he stated the question incorrectly. Instead, it should read as follows:

Show that $\sum a_n$ converges $\iff$ $\sum f(a_n)$ converges for $0 < a_n < r$, where r is sufficiently small and $f(x)=sin(x)$.

Does clear up the ongoing confusion? I am very sorry for putting forth an erred problem.

kathystehl
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  • You'll never prove this for all $f$. What if $a_n = 1$ for all $n$ and $f=0$? For $\sin$, however, notice that if $a_n \to 0$, $\sin (a_n) \eq a_n$ – hHhh Jun 23 '15 at 22:46
  • These two counterexamples show that neither implication holds [without further restriction on what types of functions $f$ you are allowed to use]: $a_n:=1/n$ and $f(x)=x^2$, as well as $a_n:=1/n^2$ and $f(x)=\sqrt{x}$. – angryavian Jun 23 '15 at 22:48
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    This doesn't work for any function $f$, but it does work for $f = \sin$. You need to look at how the function behaves close to $0$ (remember a series $\sum a_n$ converges depending roughly on how fast $a_n$ converges to $0$). – Joel Cohen Jun 23 '15 at 22:49
  • And even for $\sin$ the equivalence is wrong. You have to impose that $a_n$ converges to $0$. – hHhh Jun 23 '15 at 22:51
  • @hHhh : Ah yes, you're right, I was assuming that ! – Joel Cohen Jun 23 '15 at 22:58
  • Could someone walk me through the steps for the case $f(x)=sin(x)$? Is it always the case that $a_n$ must converge to $0$, or is it only for the case that we are assuming $f(x)=sin(x)$? I'm adding a note above, that this does not hold for all functions. – kathystehl Jun 23 '15 at 23:02
  • You can get this from $(2/\pi)x< \sin x < x, x\in (0,\pi/2). $ – zhw. Jun 23 '15 at 23:43
  • I am going to repost the question so that it makes more sense. – kathystehl Jun 24 '15 at 01:17
  • Hi all, please note the edit above. The professor gave the example incorrectly. I've included a corrected statement above. – kathystehl Jun 24 '15 at 20:10
  • @hHhh Does $a_n$ have to converge to zero, or could it converge to any value of the form $k*pi$ for $k \geq 0$, i.e. any $x$ s.t. $sin(x)=0$? – kathystehl Jun 24 '15 at 22:03
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    Wait, it won't because then $\sum a_n$ would diverge. Never mind! It has to converge to 0. – kathystehl Jun 24 '15 at 22:07

1 Answers1

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Let $f:(0,\infty) \to \mathbb {R}.$ Then $\sum f(a_n)$ converges for every positive series $\sum a_n$ that converges iff $f(x)/x$ is bounded on $(0,a)$ for some $a>0.$

A proof can be fashioned as in my post in the thread $f$ such that $\sum a_n$ converges $\implies \sum f(a_n)$ converges

(Note that there I was proposing the above as a start in solving a harder problem. Here things are easier because $\sum a_n$ is a positive series.)

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zhw.
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