$\prod (1+a_n)$ converges iff $ \sum a_n$ converges. For which range of $a_n$ it will be true. $a_n \in \Bbb R$ $\forall n \in \Bbb N$. Can anyone tell me along with the proof. I think for $a_n > 0$ it will be true but can't get proof.
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If $a_n>0$, the convergence of $\sum a_n$ is equivalent to that of $\sum \log (1+a_n) = \log \prod (1+a_n)$ – Kitegi Sep 08 '15 at 13:46
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i know that what i am saying that is there any relation in bw these two in the question. – Ri-Li Sep 08 '15 at 13:50
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I'm not sure I understand what you're asking – Kitegi Sep 08 '15 at 13:56
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Here's a proof for one direction, assuming that $a_n > 0 \ \forall n$. Let $\sum_n a_n = c$. Then, using the inequality $\log(1 + x) \le x$, and the fact that $e^x$ is monotonically increasing:
$$ \prod_n (1 + a_n) = e^{\log\prod_n(1 +a_n)} = e^{\sum_n \log(1 +a_n)} \le e^{\sum_n a_n} = e^c < \infty $$
lum
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1For the other direction, you can use the fact that $x\leq 2\log(1+x)$ for small enough $x>0$. In particular, it's true for all $0<x<2$ – Kitegi Sep 08 '15 at 14:01