Show the series an/(1-an) converges given that series an converges

Given that $$0\le a_n\lt 1$$ the series $$\sum_{n=0}^{\infty} (a_n)$$ converges. Show that the series $$\sum_{n=0}^{\infty} \frac{a_n}{1-a_n}$$ converges.

This question is supposed to be solved from first principles (e.g. Comparison Test); however, any approach would be appreciated.

I noted that $$\frac{a_n}{1-a_n} = a_n + a_n^2 + a_n^3 + ...$$ but I can't seem to finish the proof rigorously.

• When you say "series $(a_n)$ converges", do you mean "the sequence $a_1, a_2, \dots$ converges" or do you mean "the series $\sum_i a_i$ converges"? – Eric Towers May 27 at 0:22

There are only a finite number of the $$a_n$$ such that $$a_n > \frac12$$.

For all the others, $$\dfrac{a_n}{1-a_n} \lt 2a_n$$.

Since $$a_n\geq0$$ and $$\sum_na_n$$ converges, $$0\leq a_n<\frac12$$ for all $$n$$ sufficiently large. Then $$1-a_n\geq\frac12$$ and so $$0\leq\frac{a_n}{1-a_n}\leq 2 a_n$$ for all sufficiently large $$N$$. The conclusion should be easy from here. Incidentally, from your observation $$\frac{a_n}{1-a_n}=a_n+a^2_n+\ldots \geq a_n$$ It follows that if $$a_n\geq0$$, $$\sum_na_n$$ converges if and only if $$\sum_n\frac{a_n}{1-a_n}$$ converges.

$$\sum a_n\text{ converges } \implies \lim_{n\to+\infty} a_n=0$$

$$\implies \lim_{n\to+\infty} \frac{1}{1-a_n}=1$$

$$\implies \exists N : \forall n\ge N\; 0\le \frac{1}{1-a_n} \le 2$$

$$\implies \exists N : \forall n\ge N \; 0\le \frac{a_n}{1-a_n}<2a_n$$

nearly done.