Show that if $\sum_{n=1}^{\infty}a_{n}$ converges, then so do $\sum_{n=1}^{\infty}a^{2}_{n}$ and $\sum_{n=1}^{\infty}a_{n}/(1-a_{n})$.

Question

Suppose that $0 < a_{n} < 1$ for $n\in\textbf{N}$. Show that if $\sum_{n=1}^{\infty}a_{n}$ converges, then so do $\sum_{n=1}^{\infty}a^{2}_{n}$ and $\sum_{n=1}^{\infty}a_{n}/(1-a_{n})$. Are the converse statements true?

My solution

Since $0 < a_{n} < 1$, we conclude that $0 < a^{2}_{n} < a_{n} < 1$. Based on such fact, it results that $\sum_{n=1}^{\infty}a^{2}_{n}$ converges due to the comparison test. On the other hand, I do not know how to approach the second series.

Answer 1

Your proof of the first claim is correct.

If $\sum_n a_n$ converges and $a_n>0$, there is an $N\in \mathbb{N}$ such that for $n\geq N$, $a_n<1/2$. Then we could bound $a_n/(1-a_n)$ by $2a_n$ after this $N$, and use a comparison test.

Regarding the converses: the first claim is false, as we can see by taking $a_n=1/(2n+1)^2$. The second claim is true: since $\sum_n a_n/(1-a_n)$ converges and $0<a_n<1$, $a_n$ achieves a maximum value $M<1$. Then we can compare $\sum_n a_n$ with $(1-M)^{-1}\sum_n a_n$, which converges.

Answer 2

Since $0<a_n<1, \frac{a_n}{1-a_n}>0$, rewrite the summand as $e^{\log \frac{a_n}{1-a_n}}$: $$ e^{\log \frac{a_n}{1-a_n}} = e^{\log a_n - \log (1-a_n)} \approx e^{\log a_n + a_n} \leq ea_n $$ So by comaprison test the first sum converges. The $\log$ expansion is due to $a_n \to 0$, and the last bound is due to $a_n <1$.