Could you please give me some hint how to prove this statement:

If $0<a_n<1$ for each n,

then $\sum_{n=1}^\infty \frac{a_n}{1+a_n}$ converges iif $\sum_{n=1}^\infty\frac{a_n}{1-a_n}$ converges.

Since $0<\frac{a_n}{1+a_n}\le \frac{a_n}{1-a_n}$,then from convergence of $\sum_{n=1}^\infty\frac{a_n}{1-a_n}$ we may conclude by comparison test convergence of $\sum_{n=1}^\infty \frac{a_n}{1+a_n}$.

But I failed to understand how conclude from convergence of $\sum_{n=1}^\infty \frac{a_n}{1+a_n}$ convergence of $\sum_{n=1}^\infty \frac{a_n}{1-a_n}$.


  • $\begingroup$ one direction is clear, for the other if $\sum\frac{a_n}{1+a_n}$ converges, then $\sum a_n$ converges, so $a_n$ goes to zero, for large $n$ $\frac{a_n}{1-a_n}\le 2a_n$ $\endgroup$ – derivative Apr 24 '14 at 15:05
  • $\Leftarrow:0<a_n<1\implies\dfrac{a_n}{1+a_n}<\dfrac{a_n}{1-a_n}~\forall \ n$ (Use Comparison Test of the First Type)

  • $\Rightarrow:\dfrac{\dfrac{a_n}{1+a_n}}{\dfrac{a_n}{1-a_n}}=\dfrac{1-a_n}{1+a_n}=\dfrac{1-a_n}{1+a_n}=1-\dfrac{2a_n}{1+a_n}\to1~(\ne0)$ as $n\to\infty.$

    Thus ${\dfrac{a_n}{1+a_n}},{\dfrac{a_n}{1-a_n}}$ converges or diverges together. (Use Limit Form)

  • $\begingroup$ Nice! but in this case no need for the first part :) $\endgroup$ – aflous Apr 24 '14 at 15:20
  • $\begingroup$ @aflous: First part doesn't follow from the second unless the convergence of $\dfrac{a_n}{1+a_n}$ is given. $\endgroup$ – Sugata Adhya Apr 24 '14 at 16:03

To prove convergence of $\displaystyle\sum_{n=1}^\infty\frac{a_n}{1-a_n}$ given convergence of $\displaystyle\sum_{n=1}^\infty\frac{a_n}{1+a_n},$ consider using the limit comparison test. First think about what you know about $\displaystyle\lim_{n\to\infty}a_n.$ (Hint: for all $n,$ $\dfrac{a_n}{1+a_n}>\dfrac{a_n}{2}>0.)$


Notice that given the convergence of $\displaystyle\sum_{n=1}^\infty\frac{a_n}{1+a_n},$ you can prove that $a_n$ goes to $0$ as n goes to $\infty$ $$(\forall n,\dfrac{a_n}{1+a_n}>\dfrac{a_n}{2}>0)$$

and since $a_n$ is bounded from above by $1$, you could find a bound $\alpha>0$ such that: $$\forall \ n, \ 1-a_n > \alpha$$

Now write $$\frac{a_n}{1-a_n}=\frac{a_n}{1+a_n} \frac{1+a_n}{1-a_n}<2 \alpha \frac{a_n}{1+a_n}$$

The result follows

  • $\begingroup$ I'm not following you. What about $a_n = 1 - \frac{1}{n}$? It's bounded from above by $1$, but there's no such $\alpha$... $\endgroup$ – Najib Idrissi Apr 24 '14 at 15:04
  • $\begingroup$ Yes!! I rectified that. thx $\endgroup$ – aflous Apr 24 '14 at 15:18

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