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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.

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    $\begingroup$ 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"? $\endgroup$ – Eric Towers May 27 at 0:22
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There are only a finite number of the $a_n$ such that $a_n > \frac12$.

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

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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.

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

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

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

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

nearly done.

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