If $\sum\limits_{n=1}^∞|a_n|^2$ converges, then $\prod\limits_{n=1}^∞(1+a_n)$ converges if and only if $\sum\limits_{n=1}^∞a_n$ converges.

Question

If $\sum\limits_{n=1}^{\infty}|a_n|^2$ converges, then $\prod\limits_{n=1}^{\infty}(1+a_n)$ converges if and only if $\sum\limits_{n=1}^{\infty} a_n$ converges.

My attempt is to use the Cauchy citerion, given $m>n$, $$\left|\prod_{i=1}^{m}(1+a_i)-\prod_{i=1}^{n}(1+a_i)\right|=\left|\prod_{i=1}^{n}(1+a_i)\left[\prod_{i=n+1}^{m}(1+a_i)-1\right]\right|\\ =\left|\prod_{i=1}^{n}(1+a_i)\right|\left|\prod_{i=n+1}^{m}(1+a_i)-1\right|,\\ \left|\prod_{i=1}^{n}(1+a_i)\right| \leqslant \prod_{i=1}^{n}(1+|a_i|),$$ then how to proceed?

Answer 1

We first assume that $a_n\in \mathbb{R}$.

Note that since $\sum_{n=1}^\infty a_n^2$ converges, then $\displaystyle \lim_{n\to \infty}a_n=0$. There exists, therefore, a number $N$ so that $1+a_n>0$ whenever $n>N$.

Applying the Mean Value Theorem, it is straightforward to show that

$$x-\frac12x^2+\frac13x^3\le \log(1+x)\le x \tag 1$$

Applying $(1)$, we have for $n>N$

$$a_n-\frac12 a_n^2+\frac13 a_n^3 \le \log(1+a_n)\le a_n\le \frac12 a_n^2-\frac13a_n^3+\log(1+a_n) \tag 2$$ Since $\sum_{n=1}^\infty a_n^2$ converges, then so does $\sum_{n=1}^\infty a_n^3$. From $(2)$, it is easy to see that $\sum_{n=1}^\infty \log(1+a_n)$ converges if and only if $\sum_{n=1}^\infty a_n$ converges.

Finally, we note that for $n>N$

$$\prod_{n=N}^\infty (1+a_n)=e^{\sum_{n=N}^\infty \log(1+a_n)} \tag 3$$

whence we arrive at the coveted result!

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If $a_n\in \mathbb{C}$, and if $n$ is so large that $\text{Re}(1+a_n)> 0$, then $\log(1+a_n)$ can be expanded as

$$\log(1+a_n)=a_n-\frac12 a_n^2+O(a_n^3)$$

where we are free to choose to use the principal branch of the logarithm function to define $\log(1+a_n)$.

Since $\sum_{n=1}^\infty a_n^2$ converges, then so does $\sum_{n=1}^\infty a_n^3$.

Now we can proceed along similar lines as to the proof for real-valued $a_n$. The details are left as an exercise.

Answer 2

For $x\gt-1$, writing $x-\log(1+x)=\int_0^x\frac{t}{1+t}\,\mathrm{d}t$, it follows that $$ 0\le x-\log(1+x)\le\frac{\frac12x^2}{\min(1,1+x)}\tag1 $$ Thus, for $|x|\le\frac12$, we have $$ |\,x-\log(1+x)\,|\le x^2\tag2 $$ If either $\sum\limits_{n=1}^\infty a_n$ or $\sum\limits_{n=1}^\infty\log(1+a_n)$ converges, then $\lim\limits_{n\to\infty}a_n=0$. Thus, there is an $N$ so that $n\ge N\implies|a_n|\le\frac12$. Therefore, $$ \begin{align} &\left|\,\sum_{n=1}^\infty a_n-\sum_{n=1}^\infty\log(1+a_n)\,\right|\\ &\le\left|\,\sum_{n=1}^{N-1}a_n-\sum_{n=1}^{N-1}\log(1+a_n)\,\right|+\left|\,\sum_{n=N}^\infty a_n-\sum_{n=N}^\infty\log(1+a_n)\,\right|\tag{3a}\\ &\le\left|\,\sum_{n=1}^{N-1}a_n-\sum_{n=1}^{N-1}\log(1+a_n)\,\right|+\sum_{n=N}^\infty a_n^2\tag{3b}\\[9pt] &\lt\infty\tag{3c} \end{align} $$ Explanation:
$\text{(3a)}$: triangle inequality
$\text{(3b)}$: apply $(2)$
$\text{(3c)}$: the sum of two finite sums is finite