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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 $a_n=-1$.

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Should that be $\prod(1+a_n)$ in the title? –  Brian M. Scott Apr 9 '12 at 2:39
    
@BrianM.Scott Yup, thanks! –  Steven-Owen Apr 9 '12 at 2:40
    
Related: math.stackexchange.com/questions/119140/… –  Aryabhata Apr 9 '12 at 4:34

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up vote 3 down vote accepted

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.

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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 9 '12 at 3:37
    
@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 9 '12 at 4:12
    
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 9 '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 9 '12 at 18:11

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