# a divergent series which $\prod 1+a_n$ converges.

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

• If $a_n$ is allowed to be negative, then there are such sequences $\{a_n\}$. Should we assume $a_n>0$? – alex.jordan May 16 '14 at 5:14
• Harmonic series? – chubakueno May 16 '14 at 5:15
• @chubakueno The harmonic series here would not provide a convergent product. – alex.jordan May 16 '14 at 5:15
• @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? – chubakueno May 16 '14 at 5:19
• @user115608 Well, it depends if you consider a limit of $0$ to make for a convergent product. See Bruno's answer. – 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.

• 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. – BlackAdder May 16 '14 at 5:55
• 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. – Bruno Joyal May 16 '14 at 5:57
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.