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

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

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.

  • $\begingroup$ 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. $\endgroup$ – alpha.Debi Apr 14 '12 at 13:31
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    $\begingroup$ @alpha.Debi $${a_n\over 1+a_n}<{1\over2}\ \Longrightarrow\ a_n<{1\over2}+{a_n\over2}\ \Longrightarrow\ {a_n\over2}<{1\over2}\ \Longrightarrow\ {a_n<1}.$$ $\endgroup$ – David Mitra Apr 14 '12 at 13:34
  • $\begingroup$ Nice proof (all the proof) , thanks. $\endgroup$ – alpha.Debi Apr 14 '12 at 13:38
  • $\begingroup$ Thank you for your answer, your answer is always nice!!! $\endgroup$ – Riemann Apr 14 '12 at 14:06
  • $\begingroup$ 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. $\endgroup$ – Pedro Tamaroff Jan 27 '15 at 23:25

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.)


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.


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


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.

  • $\begingroup$ We can restrict to the last two cases but I started with the most common to think about. $\endgroup$ – alpha.Debi Apr 14 '12 at 12:14

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