I'm aware that similar questions have been posted before, for example here. But my question is not a duplicate since I am asking about my specific proof. Is it correct, and could it have been made simpler while retaining the same basic idea?

If $a_n\not\to 0$, let $\epsilon > 0$ be such that $a_n > \epsilon$ for infinitely many values of $n$. Then $$\frac{a_n}{1+a_n} = \frac{1}{1/a_n+1} > \frac{1}{\epsilon+1}$$ for infinitely many values of $n$. Thus $\frac{a_n}{1+a_n}\not\to 0$, and the series diverges.

On the other hand, assume that $a_n\to 0$. If $\sum\frac{a_n}{1+a_n}$ converges, then so does $\sum\frac{a_n^2}{1+a_n}$, by the comparison test. But then $$\sum a_n = \sum\left(\frac{a_n}{1+a_n} + \frac{a_n^2}{1+a_n}\right) = \sum\frac{a_n}{1+a_n} + \sum\frac{a_n^2}{1+a_n}$$ converges, a contradiction.

Source: This is problem 11(a) in chapter 3 of Rudin's Principles of Mathematical Analysis.

  • 2
    $\begingroup$ It needs to be $\frac{a_n}{1+a_n}>\frac{\epsilon}{1+\epsilon}$, but everything else looks fine. $\endgroup$ – Mark Jan 15 at 19:35

Yes, your arguments are correct, with the only caveat that your first inequality should have been $$ \frac{a_n}{a_n+1}>\frac1{\frac1\epsilon+1}. $$

  • 1
    $\begingroup$ Wow, we posted the same thing at the same time. $\endgroup$ – Mark Jan 15 at 19:35
  • $\begingroup$ Thanks to both of you for the response. $\endgroup$ – Brennan Vincent Jan 15 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.