A question from Rudin (Principles) Chapter 3:

Let $a_n\geq0$ and $\sum a_n$ diverges. What can be said about convergence/divergence of $\sum\frac{a_n}{1+na_n}$?

This one is being recalcitrant. Given that $x>y$ implies $\frac{x}{1+nx}>\frac{y}{1+ny}$ and when $a_n=1/n\log n$ the sum in question diverges, it seems plausible that in general the sum will always diverge, but I can't get a proof out. If it does diverge, it does so pretty slowly as $$\frac{a_n}{1+na_n}=\frac{1}{n}-\frac{1}{n+n^2a_n}\leq\frac{1}{n}.$$

  • $\begingroup$ I'd try some example divergent sequences first. $a_n=1$ the new series does not converge. Same for $a_n=\frac{1}{n}$. $\endgroup$ – Thomas Andrews Apr 11 '12 at 15:38

Let $A=\sum\limits_n a_n$ denote a series with nonnegative terms, and $B=\sum\limits_nb_n$ with $b_n=\frac{a_n}{1+na_n}$. Of course, $b_n\leqslant a_n$ hence, if $A$ converges, so does $B$. But, if $A$ diverges, $B$ may diverge or converge.

Example 1: Assume that $a_n=\frac1n$ for every $n\geqslant1$. Then, $b_n=\frac1{2n}$ for every $n\geqslant1$ hence $A$ and $B$ both diverge.

Example 2: Assume that $a_n=1$ for every $n$ in some subset $S$ of the integers, and $a_n=0$ otherwise. Then, $A$ diverges if and only if $S$ is infinite and $B$ diverges if and only if $\sum\limits_{n\in S}\frac1{n+1}$ diverges. If $S$ is the set of squares of integers, $A$ diverges and $B$ converges.

Example 2': If one needs that every $a_n$ is positive, assume that $a_n=1$ for every $n$ in $S$ the set of squares of integers and $a_n=\frac1{n^2}$ otherwise. Then, $A$ diverges and $B$ converges.

  • $\begingroup$ +1: Very strange! I was about to add an answer with your exact Example 2. Find an infinite set of positive integers the sum of whose reciprocals converges. Take $a_n$ to be the indicator sequence for that set. $\endgroup$ – Aryabhata Apr 11 '12 at 16:19
  • $\begingroup$ @Aryabhata Thanks, les grands esprits se rencontrent, as they say... :-) And if $(a_n)$ is nonincreasing, what do you think? $\endgroup$ – Did Apr 11 '12 at 16:23
  • $\begingroup$ Haven't thought about it much, but for $B$ to converge, we must have $a_n \to 0$. $\endgroup$ – Aryabhata Apr 11 '12 at 16:31

It is easy to produce divergent $\sum a_n$ such that $\sum\frac{a_n}{1+a_n}$ diverges. But convergence is also possible, as in the following example.

Let $a_1=1$, $a_2=\frac{1}{2}$, $a_3=0$, $a_4=\frac{1}{3}+\frac{1}{4}$, $a_5=a_6=a_7=0$, $a_8=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}$, $a_9=a_{10}=\cdots=a_{15}=0$, $a_{16}=\frac{1}{9}+\frac{1}{10}+\cdots+\frac{1}{16}$, and so on.

Then the series $\sum a_n$ diverges, it is essentially the harmonic series.

But the series whose $n$-th term is $\frac{a_n}{1+na_n}$ converges, it behaves for all practical purposes like a geometric series.

Remark: More simply, we can replace $\frac{1}{2}$ by $1$, $\frac{1}{3}+\frac{1}{4}$ by $1$, $\frac{1}{5}+\cdots+\frac{1}{8}$ by $1$, and so on. But (i) the example in the post was the first one I thought of, and (ii) the idea can be used to "mess up" any divergent series $\sum a_n$ with non-negative terms that approach $0$.


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