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Let $(a_n)_n$ be a sequence of non-negative real numbers.

Prove that $\sum\limits_{n=1}^{\infty}a_n$ converges if and only if the infinite product $\prod\limits_{n=1}^{\infty}(1+a_n)$ converges.

I don't necessarily need a full solution, just a clue on how to proceed would be appreciated.

Could any convergence test like the ratio or the root test help?

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Hint: Since the $a_n$ are nonnegative, we have the inequalities

$$\sum_{n=1}^N a_n \le \prod_{n = 1}^N (1 + a_n) \le \exp\left\{\sum_{n = 1}^N a_n\right\}\quad (N = 1,2,\ldots)$$

where the second inequality follows from the inequality $1 + x \le e^x$ for all $x \ge 0$.

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    $\begingroup$ Thanks! I guess one should add "0≤" at the left end of the chain of inequalities because the series must be non-negative in order to rule out divergence to -∞. $\endgroup$ – Mophotla Nov 23 '16 at 18:51
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Hint: Consider $$ \ln\left[\prod_{n=1}^\infty (1+a_n) \right] $$ and note that if $a_n \to 0$, the Taylor series for $\log(1+x)$ is a useful approximation.

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A way is to use logarithm and the equivalence $$\ln(1+x) \sim x$$ around the origin.

Interesting to notice is that for complex series (or series not having a constant sign) $\sum z_n$, the absolute convergence of the series implies the convergence of the product $\displaystyle \prod (1+z_n)$, but the converse is not always true. See my counterexamples web site.

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