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Find a sequence $\{a_n\}$ of real numbers such that $\sum a_n$ converges but $\prod (1+ a_n)$ diverges.

The converse is trivial, just make all the $a_n=-1$.

Steven-Owen
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1 Answers1

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It's not hard to show that if each $|a_n| < 1$ then $\prod_n(1 + a_n)$ converges to a nonzero value iff $\sum_n \log(1 + a_n)$ does (take logarithms of the partial products). If $a_n \rightarrow 0$ then for $n$ large enough, by Taylor expanding the log you have for example $$ a_n -a_n^2 < \log(1 + a_n) < a_n - {a_n^2 \over 4}$$ So if you take $a_n = (-1)^n{1 \over \sqrt{n}}$, even though $\sum {a_n}$ converges (it's a decreasing alternating series), the sum of the logarithms will diverge since $a_n^2 = {1 \over n}$. So the product will diverge as well.

Zarrax
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  • Hello, first of $a_n-a_n^2 \log(1+a_n)$ is only true for $a_n >0$ which is not our case if we take an alternating series. Also, isn't it difficult to show that $\sum (-1)^n\frac1{\sqrt{n}}-1/n$ diverges. – Steven-Owen Apr 09 '12 at 03:37
  • @jake $\log(1 + a_n) = a_n - {a_n^2 \over 2} + o(a_n^2)$ regardless if $a_n$ is positive. And yes, that sum diverges to $-\infty$, that's the point. So the product goes to zero. (It will never go to infinity). – Zarrax Apr 09 '12 at 04:12
  • Sorry, I still don't understand, should it be $log(1+a_n)=a_n-\frac{a_n}2+o(a_n^3)$ where $o(a_n^3)$ will be negative if $a_n^3$ is negative so the inequality doesn't hold. – Steven-Owen Apr 09 '12 at 15:41
  • Also I still don't see how you conclude that $\sum (-1)^n\frac1{\sqrt{n}}-\frac1n$ diverges. Every convergence test I've done is inconclusive. – Steven-Owen Apr 09 '12 at 18:11
  • @Zarrax The product can still converge to zero right?? I computed a bit using SageMath and found that the above product goes close to zero as $n$ becomes large. So I don' t see how your answer is complete. – Sundara Narasimhan May 20 '20 at 14:56
  • @S.SundaraNarasimhan The usual definition of convergence for an infinite product requires the limit to be nonzero. – Zarrax May 20 '20 at 22:02
  • @Zarrax If that is the case it is OK, also I'm curious if there are any such sequences that don't converge to any number. – Sundara Narasimhan May 21 '20 at 07:48
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    For simplicity assume each $|a_n| < 1$. Because of the inequality $ a_n -a_n^2 < \log(1 + a_n) < a_n - {a_n^2 \over 4}$, when $\sum_n a_n$ converges the sum of the $ \log(1 + a_n) $ either converges to a finite number (if $\sum_n a_n^2$ is finite) or equals $-\infty$ (if $\sum_n a_n^2$ is infinite) . In the first case the product converges to a nonzero number and in the second case it goes to zero. But always one of the two cases will hold. – Zarrax May 23 '20 at 18:15
  • The inequality a-a^2<log(1+a) does not hold for |a|<1. For example for a=1/2, we have a-a^2=1/4, but log(3/2)=0.176091...

    Taking |a| smaller doesn't help either, as can be seen by graphing log(1+x)-x+x^2

     – TheEmptyFunction Nov 03 '22 at 22:47
  • @TheEmptyFunction I should have said for sufficiently small $|a|$. If you graph $\log(1 + x) - x + x^2$ on say $[-0.1,0.1]$, then you can see the function is positive. – Zarrax Nov 04 '22 at 01:55
  • Ah, whoops! The graphing software I was using interprets log as the base 10 logarithm. That's why the inequality wasn't holding – TheEmptyFunction Nov 04 '22 at 17:52