i am searching for a series with this condition that $\prod 1+a_n$ converges but $\Sigma a_n$ diverges.

i know that if $a_n = n^{\frac{1}{2}}$ then $\Sigma a_n$ diverges but i dont know it is exactly what i want, does $\prod 1+a_n$ converges?

i really don't know how to check the divergence or convergence of a product series.

if i'm wrong can anyone tell me such example?

thank u

  • 1
    $\begingroup$ If $a_n$ is allowed to be negative, then there are such sequences $\{a_n\}$. Should we assume $a_n>0$? $\endgroup$ – alex.jordan May 16 '14 at 5:14
  • $\begingroup$ Harmonic series? $\endgroup$ – chubakueno May 16 '14 at 5:15
  • $\begingroup$ @chubakueno The harmonic series here would not provide a convergent product. $\endgroup$ – alex.jordan May 16 '14 at 5:15
  • $\begingroup$ @alex.jordan Actually it was really a question, I wasn't really sure :) May I know why? I see a telescopic product whose $n$th term goes to $\frac{n+1}{n}$ . Maybe this doesn't fit the usual definition of convergent product, I don't know. So, may you enlighten me? $\endgroup$ – chubakueno May 16 '14 at 5:19
  • 1
    $\begingroup$ @user115608 Well, it depends if you consider a limit of $0$ to make for a convergent product. See Bruno's answer. $\endgroup$ – alex.jordan May 16 '14 at 5:22

The usual definition of convergence for infinite products rules out such a possibility by definition.

In any case, if the $a_n\geq 0$, then the product converges to a finite limit if and only if $\sum a_n<\infty$.

If you consider a product which converges to $0$ to be convergent, then there are examples. For instance $(1-1/2)(1-1/3)\dots$ "converges to $0$", but it is usually considered to be a divergent product for this very reason. Even sillier: $(1-1)(1-1)\dots = 0$ but $1+1+\dots$ diverges.

  • $\begingroup$ Hi, I would just like to point out that the wikipedia article you reference has the following definition: The product converges if and only if $\sum \log a_n$ converges, not $\sum a_n$. Just a pedantic point. $\endgroup$ – BlackAdder May 16 '14 at 5:55
  • 3
    $\begingroup$ Dear @BlackAdder: The product considered here is $\prod (1+a_n)$, not $\prod a_n$. The statement I have written is correct, as you can read further in the wikipedia article. $\endgroup$ – Bruno Joyal May 16 '14 at 5:57
  • $\begingroup$ I misread. Sorry! $\endgroup$ – BlackAdder May 16 '14 at 6:23
  • $\begingroup$ @BlackAdder No worries! $\endgroup$ – Bruno Joyal May 16 '14 at 17:08

According to the usual definition (as mentioned by Bruno), "convergence" for an infinite product means "convergence of the partial products to a nonzero limit". So, assuming $a_n>-1$ for all $n$, the convergence of $\prod (1+a_n)$ is equivalent to the convergence of the series $\sum\log(1+a_n)$.

Hence, the question is: to find a sequence $(a_n)$ such that the series $\sum a_n$ is divergent but the series $\sum\log(1+a_n)$ is convergent.

Consider the sequence defined by $$a_n=\frac{(-1)^n}{\sqrt n}+\frac1{2n}\cdot $$ Then $\sum a_n$ is divergent because $\sum\frac{(-1)^n}{\sqrt n}$ is convergent and $\sum\frac1n$ is divergent. Let us show that, on the other hand, the series $\sum\log(1+a_n)$ is convergent.

By the Taylor expansion for $\log(1+u)$, we may write $$\log(1+a_n)=a_n-\frac{a_n^2}2+O(a_n^3)\, . $$ Since $\vert a_n\vert\sim\frac1{\sqrt n}$, the $O(a_n^3)$ term is $O(1/n^{3/2})$ and hence the corresponding series is convergent. So it is enough to show that the series $\sum(a_n-\frac{a_n^2}2)$ is convergent. Now we have \begin{eqnarray}a_n-\frac{a_n^2}2&=&\frac{(-1)^n}{\sqrt n}+\frac1{2n}-\frac12\left(\frac1n+\frac{(-1)^n}{n^{3/2}}+\frac1{4n^2} \right) \\&=&\frac{(-1)^n}{\sqrt n}+O\left(\frac1{n^{3/2}}\right) , \end{eqnarray} so the series is indeed convergent, being the sum of a convergent alternating series and an absolutely convergent series.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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