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 $a_n=-1$.
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 $a_n=-1$.
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.