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

Let $\sum\limits_{n\geq1}a_n$ be a positive series, and $\sum\limits_{n\geq1}a_n=+\infty$, prove that: $$\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty.$$

share|cite|improve this question
Please try to add in the question as to what progress you have made. People here will help you build up on that. – user9413 Apr 14 '12 at 11:20
This is a question in the text book in Rudin's Principles of Mathematical Analysis. – Riemann Apr 14 '12 at 11:29
fantastic. thanks for referring to the book. – vondip May 18 '13 at 11:18
up vote 18 down vote accepted

Proving the contrapositive statement seems cleaner to me. Suppose $\sum{a_n\over 1+{a_n}}$ converges. Then ${a_n\over 1+{a_n}}\rightarrow 0$. This implies that ${a_n}$ is eventually less than one, so ${a_n\over2}\le {a_n\over a_n+1}$ for $n$ sufficiently large. The comparision test then shows that $\sum a_n$ converges.

share|cite|improve this answer
How do you prove a(n) is less than one ? If a(n) is bounded you can multiply and divide a(n) by 1+a(n) and you get bounded multiplied by a factor which converges to zero, then a(n) converges to zero ad you continue in that way ypu wrote. – alpha.Debi Apr 14 '12 at 13:31
@alpha.Debi $${a_n\over 1+a_n}<{1\over2}\ \Longrightarrow\ a_n<{1\over2}+{a_n\over2}\ \Longrightarrow\ {a_n\over2}<{1\over2}\ \Longrightarrow\ {a_n<1}.$$ – David Mitra Apr 14 '12 at 13:34
Nice proof (all the proof) , thanks. – alpha.Debi Apr 14 '12 at 13:38
Thank you for your answer, your answer is always nice!!! – Riemann Apr 14 '12 at 14:06
David, apologies in advance for changing the colour of $a_n$; but maroon is dark and close to black and it seems to upset the eyes of some people with colour issues. – Pedro Tamaroff Jan 27 '15 at 23:25

Alternatively, split the problem up in cases:

1, If there is a natural $N\in\mathbb{N}$ s.t $a_n\leq{1}$ for $n\geq N$, what can you conclude?

2, If (1) is not true, for every natural $N$ we can find a $n\geq{N}$ s.t $a_n>1.$ Now passing to a subsequence and comparing with a series with each term equal to a constant (more precisely $1/2$ or lower), the result follows.

share|cite|improve this answer

I would argue by cases.

Case 1. $a_n\ge1$ for infinitely many $n\in\mathbb N$.

In this case, for each such $n$ we have $\frac{a_n}{1+a_n}=1-\frac{1}{1+a_n}\ge1-\frac12$, from which the claim easily follows.

Case 2. $a_n\ge1$ for only finitely many $n\in\mathbb N$.

In this case for every other $n$ we have $a_n<1$ and thus $\frac{a_n}{1+a_n}\ge\frac{a_n}{1+1}=\frac{a_n}2$. Since finitely many terms can't affect the convergence/divergence of a series, this will also diverge. (Since $\sum\limits_{n=1}^\infty\frac{a_n}{2}$ does.)

share|cite|improve this answer

You can suppose that $(a_n) \rightarrow 0$ (if not the problem is trivial). Then what can you say asymptotically about $\frac{a_n}{1+a_n}$ ?

share|cite|improve this answer

We can divide into cases:

  1. If a(n) has limit zero : It is lower than 1 for all n bigger than n0, then we can compare with a(n)/2 which is lower than a(n)/(1+a(n)).

  2. If a(n) has limit different to zero , also a(n)/1+a(n) and then the series diverges

  3. If a(n) is not bounded it ha a subsequence that converges to infinite, then a(n)/1+a(n) converges to 1 then the series diverges to infinite.

  4. If a(n) is bounded , we can take a subsequence that is convergent.

If it does not converges to zero also the sequence a(n)/1+a(n). If all subsequences converge to zero ,then also a(n) and we can apply 1.

share|cite|improve this answer
We can restrict to the last two cases but I started with the most common to think about. – alpha.Debi Apr 14 '12 at 12:14

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.