$a_n$ is an infinite sequence with $\sum\limits_{n=1}^\infty a_n\leq1$ and $0\leq a_n<1$. Prove that $\sum\limits_{n=1}^\infty a_n/(a_n-1)$ converges?

Question

For every infinite sequence that sums to one that I can think of, for example $a_n = 2^{-n}$, the sum $\sum\limits_{n=1}^\infty \frac{a_n}{a_n-1}$ converges, but how can I prove this? I tried the definition of a Cauchy sequence, the ratio test, but:

$$\lim\limits_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| = 1$$

so it is inconclusive, and I don't see how I can use the root test. I also don't see a convergent sequence that is below the sequence above.

Thank you for your help!

Answer 1

Let $\epsilon>0$. As $\sum a_n$ converges, we have $a_n\to 0$ and in particular $a_n<\frac \epsilon{1+\epsilon}$ for almost all $n$. Apart from the finitely many summands with $a_n\ge \frac \epsilon{1+\epsilon}$, we have $a_n<\frac{a_n}{1-a_n}<(1+\epsilon)a_n$. Then the liminf and limsup of the partial sums of $\sum\frac{a_n}{1-a_n}$ differ by at most $\epsilon\sum a_n\le \epsilon$. As $\epsilon$ was arbitrary positive, we conclude that $\liminf=\limsup$.

Answer 2

You have

• $\lim_{n\to \infty} a_n = 0$ since $\sum_{n=1}^\infty a_n \leq 1$.
• $\Rightarrow \lim_{n\to \infty} \frac{1}{a_n - 1} = -1$.
• $\Rightarrow -1 -\epsilon < \frac{1}{a_n - 1}< -1 +\epsilon$ for $n\geq N_{\epsilon}$. $$\stackrel{a_n \geq 0}{\Rightarrow} (-1-\epsilon)\sum_{n=N}^\infty a_n \leq \sum_{n=N}^\infty \frac{a_n}{a_n - 1} \leq (-1+\epsilon)\sum_{n=N}^\infty a_n$$ So, it is convergent.

Answer 3

Attempt at an answer (with help from the comments on my question):

Let me sort the $a_n$ in descending order, then $a_n < 1/2$ for $n \geq 2$, and the denominator $a_n - 1 < -1/2$, so $| \frac{a_n}{a_n -1} | \leq a_n$ and I've found a comparator series.

Answer 4

Hint: Since the sum is convergent the terms tend to $0$. So there exists $N$ such that $0\le a_n<1/2$ for all $n>N$. But $0\le a_n<1/2$ implies $1/(1-a_n)<2$.