Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $\sum a_n$ is convergent. Is $\sum {{a_n} \over {1+|a_n|}}$ convergent or divergent?

share|cite|improve this question
First consider the simple case that all $a_n\ge0$ and compare $a_n$ with $\frac{a_n}{1+|a_n|}$. – Hagen von Eitzen Sep 10 '12 at 6:23
What about the case that part of $a_n$ are positive and part of it are negative – Mathematics Sep 10 '12 at 6:54

If $\sum a_n$ converges absolutely, the the answer is affimative. We claim that this is no longer the case for conditional convergence. Note that

$$ \frac{x}{1+|x|} = x - x|x| + O(x^3)$$

near the origin. Now consider the series

$$\sum_{n=1}^{\infty} a_n = \frac{2}{\sqrt{1}} - \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{1}} + \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{3}} + \cdots.$$

This series converges conditionally. Now then we have

$$ \frac{a_{3n-2}}{1+|a_{3n-2}|} + \frac{a_{3n-1}}{1+|a_{3n-1}|} + \frac{a_{3n}}{1+|a_{3n}|} = -\frac{2}{n} + O\left( \frac{1}{n^{3/2}}\right). $$

Therefore the sum $\sum \frac{a_n}{1+|a_n|}$ diverges.

Slightly modifying this argument also generates a conditionally convergent series $\sum a_n$ whose corresponding sum $\sum \frac{a_n}{1+|a_n|}$ also converges, thus the answer is inconclusive.

share|cite|improve this answer

Surprisingly, it need not be convergent. In fact, if $f$ is a function such that $\sum_n a_n$ converges iff $\sum_n f(a_n)$ converges, then $f$ must be linear in some neighbourhood of $0$. See!topic/sci.math.research/SzZDXRFyvhk

share|cite|improve this answer
Unless I am mistaken, the main information this forum page contains is that This was proved by G. Waldenberg, American Mathematical Monthly 95 (1988) 542-544. Y. Benjamini's solution to Problem E3404, American Mathematical Monthly 99 (1992) 466-467 contains an extension. Let me suggest that you include this in your answer. – Did Sep 10 '12 at 8:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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