3
$\begingroup$

This question already has an answer here:

I'm having a look at analysis right now, and I just thought up this question after reading about the comparison test.

Does there exist a "critical" infinite sum of real numbers (which is divergent) such that if $S_n$ is any infinite sum for which the terms in $S_n$ are all less than the terms in this critical sum for $n > N_0$ for some finite $N_0$, then $S_n$ converges?

I know my terminology is somewhat incorrect.

$\endgroup$

marked as duplicate by Travis, Namaste, Eric Stucky, Ivo Terek, hardmath Jan 19 '15 at 15:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You have to add that the critical sum has to be divergent $\endgroup$ – daw Jan 19 '15 at 9:35
  • $\begingroup$ @daw: "critical infinite sum" is not good enough (assuming that the "infinite" refers to the sum and not to the number of terms in it)? $\endgroup$ – barak manos Jan 19 '15 at 9:35
  • $\begingroup$ Thanks, added, and yeah it well might be, I'm typing this on mobile on the train $\endgroup$ – Nethesis Jan 19 '15 at 9:36
  • $\begingroup$ Didn't have fast enough internet to load anything else $\endgroup$ – Nethesis Jan 19 '15 at 9:36
  • $\begingroup$ I don't think it's quite a duplicate though, as there is a difference between smallest divergent sum and largest convergent sum. It's like the difference between smallest positive real number and largest non-positive real number. $\endgroup$ – Nethesis Jan 19 '15 at 9:38
2
$\begingroup$

No, if $\sum a_n$ is divergent (positive terms) then $\sum \frac{a_n}{s_n}$ is also divergent.

Exp: $a_n = \frac{1}{n}$, $s_n \simeq \log n$.

$\bf{Added:}$ Let's try to prove it. Enough to show that $\sum_n \frac{a_n}{s_n}$ is not Cauchy. Let $n$ be arbitrary. For $N>n$ we have

$$\frac{a_{n+1}}{s_{n+1}} + \cdots + \frac{a_N}{s_N} \ge \frac{a_{n+1} + \cdots + a_{N}}{s_N} = \frac{s_N - s_n}{s_N}= 1 - \frac{s_n}{s_N} $$

and this can be made $>\frac{1}{2}$ for $N$ large enough. It follows that $\sum_n\frac{a_n}{s_n}$ is divergent.

$\endgroup$
  • $\begingroup$ Can't find a reference right now, pretty standard fact. $\endgroup$ – Orest Bucicovschi Jan 19 '15 at 9:39
  • 1
    $\begingroup$ Does that mean there is no way to quantify the "growth rate" of a series? $\endgroup$ – Nethesis Jan 19 '15 at 9:46
  • $\begingroup$ Of course, you can quantify the "growth rate" of $a_n$. No problem with that. If you want you can do that with the $s_n$. Again, given your $a_n$ with $s_n \to \infty$ you can find $b_n$ with $s_n' \to \infty$ and yet $\frac{s_n'}{s_n} \to 0$, it's not that hard. This was about the estimate for the terms of the series. $\endgroup$ – Orest Bucicovschi Jan 19 '15 at 10:13
  • $\begingroup$ @daw: check the proof I added, it's a standard result. $\endgroup$ – Orest Bucicovschi Jan 19 '15 at 10:13
  • $\begingroup$ got it, misread your answer. You should add the definition of $s_n:=\sum_{i=1}^na_n$. $\endgroup$ – daw Jan 19 '15 at 11:18

Not the answer you're looking for? Browse other questions tagged or ask your own question.