Suppose $a_n>0$ for $n\in \mathbb{N}$. Prove that $\prod_{n=1}^\infty (1+a_n)$ converges if and only if $\sum_{n=1}^\infty a_n<\infty$.

Question

Suppose $a_n>0$ for $n\in \mathbb{N}$. Prove that $$\prod_{n=1}^\infty (1+a_n)$$ converges if and only if $\sum_{n=1}^\infty a_n<\infty$.

Hint: Use the fact that for any $a>0$, $$a>\ln(1+a)>\frac{a}{1+a}.$$

Assume that $\sum_{n=1}^\infty a_n<\infty$. By the hint, $$\infty> \sum_{n=1}^\infty a_n>\sum_{n=1}^\infty \ln(1+a_n)>\sum_{n=1}^\infty \frac{a_n}{1+a_n}.$$ So $$\infty>\sum_{n=1}^\infty \ln(1+a_n).$$ By the limit comparison test $$\infty> \sum_{n=1}^\infty (1+a_n).$$ This implies that $$\prod_{n=1}^\infty (1+a_n)$$ converges.

How would I go from the other direction?

Hint: You haven't used yet that $\log(1+a_n)>\frac{a_n}{1+a_n}$. Show that if $\sum \frac{a_n}{1+a_n}$ converges then $\sum a_n$ converges. — Thomas Andrews, Feb 11 '16 at 00:22
It's somewhat sloppy, language-wise, but it is basically correct. — Thomas Andrews, Feb 11 '16 at 00:32
Related: $\sum_{n=1}^\infty a_n<\infty$ if and only if $\sum_{n=1}^\infty \frac{a_n}{1+a_n}<\infty$ — Winther, Feb 11 '16 at 00:45

score 1 · Accepted Answer · answered Feb 11 '16 at 00:23

On the other direction, notice that necessarily $a_n\to0$ because $\prod_{i=1}^\infty (1+a_n)$ converges. Hence, $$ \frac12<\frac1{1+a_n}<2 $$ for all sufficiently large $n$. So, by comparison, $\sum_{i=1}^\infty a_n$ converges if and only if $\sum_{i=1}^\infty a_n/(1+a_n)$ does (and the last one converges because $\prod_{i=1}^\infty (1+a_n)$ does).

Suppose $a_n>0$ for $n\in \mathbb{N}$. Prove that $\prod_{n=1}^\infty (1+a_n)$ converges if and only if $\sum_{n=1}^\infty a_n<\infty$.

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