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.
$\endgroup$ 13 Answers
$\begingroup$There are only a finite number of the$a_n$ such that$a_n > \frac12$.
For all the others,$\dfrac{a_n}{1-a_n} \lt 2a_n$.
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$$$\sum a_n\text{ converges } \implies \lim_{n\to+\infty} a_n=0$$
$$\implies \lim_{n\to+\infty} \frac{1}{1-a_n}=1$$
$$\implies \exists N : \forall n\ge N\; 0\le \frac{1}{1-a_n} \le 2$$
$$\implies \exists N : \forall n\ge N \; 0\le \frac{a_n}{1-a_n}<2a_n$$
nearly done.
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