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
 A: 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.
A: 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.
