Let $(a_n)_n$ be a sequence in $\mathbb{C}$, $a_n \neq -1$.

Prove that $\displaystyle\sum_{n=1}^{\infty}|a_n| < \infty,$ then $\displaystyle\prod_{n=1}^{\infty}(1+a_n)$ converges to a non-zero number.

Here is what I have so far $\displaystyle\prod_{n=1}^{\infty}|1+a_n| \leq \displaystyle\prod_{n=1}^{\infty}(1+|a_n|) \leq \displaystyle\prod_{n=1}^{\infty}e^{|a_n|} = \displaystyle\exp(\sum_{n=1}^{\infty}|a_n|) < \infty$

So I only showed that the infinite product is bounded above, but how do I show the following

  1. $\displaystyle\prod_{n=1}^{\infty}(1+a_n)$ converges. (By Monotone Convergence Theorem, we know that $\displaystyle\prod_{n=1}^{\infty}(1+|a_n|)$ converges, but I am not sure where to go from there

  2. How to show that $\displaystyle\prod_{n=1}^{\infty}|1+a_n| >0$, In the real case, this follows from $e^{-|a_n|} \leq 1+|a_n| \leq e^{|a_n|}$ But I am not sure how to generalize that to the complex norm

  • 2
    $\begingroup$ maybe using the logarithm helps $\endgroup$ Commented Oct 28, 2017 at 18:58
  • $\begingroup$ Your third line need editing. $\endgroup$
    – zhw.
    Commented Oct 28, 2017 at 19:10

1 Answer 1


Taking for granted that $\exp$ is continuous, we just need to show that $\sum_{k=1}^\infty \log (1+a_k)$ converges absolutely if $\sum_{k=1}^\infty a_k$ does. Note that the assumption implies that $a_k \to 0$ so that the (principal) logarithm is well defined for large enough $k$; without loss $k=1$ is large enough.

For this, note the (imprecise) inequality from e.g. Taylor expansion that $$ |\log(1+a_k)| ≤ 10|a_k|$$ so that $$\sum_{k=1}^K|\log(1+a_k)| ≤ 10\sum_{k=1}^K |a_k| \to 10 \sum_1^\infty |a_k| < \infty$$ This implies that $\sum_{k=1}^\infty \log (1+a_k)$ converges absolutely, and therefore the product $$ \prod_{k=1}^\infty (1+a_k) := \lim_{K\to \infty}\exp \left(\sum_{k=1}^K \log(1+a_k)\right) = \exp \left(\sum_{k=1}^\infty \log(1+a_k)\right)$$ converges.

It is a positive number because $\exp$ restricted to the reals has positive range, and $∏_{k=1}^\infty |1+a_k| = \exp (\sum_{k=1}^\infty\log|1+a_k|)$ (with the same definition of infinite product above) is the exponential of a real number.

  • $\begingroup$ Thank you, but can you explain the Taylor Expansion Inequality part a bit more? $\endgroup$
    – Phantom
    Commented Oct 28, 2017 at 20:20
  • $\begingroup$ @Phantom do you know Taylor's theorem? $\endgroup$ Commented Oct 28, 2017 at 20:39
  • $\begingroup$ You don't really need Taylor, just $\log'(1) = 1,$ which gives $$\left | \frac{\log(1+z)}{z} \right | \le 2$$ for small $z.$ This leads to $$|\log(1+a_n)|\le 2|a_n|$$ for large $n.$ $\endgroup$
    – zhw.
    Commented Oct 29, 2017 at 20:44
  • $\begingroup$ @zhw That's Taylor for a $C^1$ function, isn't it? Or am I misnaming things. $\endgroup$ Commented Oct 29, 2017 at 20:46

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