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I'm writing solutions to exercises in Baby Rudin I think might want to assign students, and I'm having particular difficulty with 11(d) in Chapter 3. Namely, for a sequence $(a_{n})_{n=1}^{\infty}$ of positive reals such that $\sum_{n=1}^{\infty}a_{n}$ diverges, does $\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ converge or diverge?

I know that if we take $a_{n} := \frac{1}{n}$, then $\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}} =\sum_{n=1}^{\infty}\frac{1}{2n}$ diverges. But I can't seem to come up with an example such that $\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ converges while $\sum_{n=1}^{\infty}a_{n}$ diverges. I did a bit of searching online, but the example I came across was $a_{n} = \frac{1}{n(\log n)^{p}}$, where $p > 1$. However, this isn't satisfactory since $\sum_{n=2}^{\infty}\frac{1}{n(\log n)^{p}}$ converges by Theorem 3.29 in Baby Rudin.

Can anyone help me out?

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See here math.stackexchange.com/questions/214556/… –  blindman Nov 15 '12 at 3:46
    
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2 Answers 2

up vote 4 down vote accepted

Take $a_{m^2} = 1/m$, while $a_n = 2^{-n}$ if $n$ is not a square. Note that $a_{m^2}/(1 + m^2 a_{m^2}) \le 1/m^2$.

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Thank you, Robert! My apologies to the community for not doing a better job searching before posting; I'm a bit embarrassed. –  Matt Nov 15 '12 at 3:56
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Same idea as Robert Israel. Let $a_n=1$ if $n=2^k$, and $a_n=e^{-n}$ otherwise.

Remark: The intuition here is that $a_n$ is $1$ on a sparse set, and $0$ otherwise. But the question specifies that the $a_n$ are positive. The fix is easy, make $a_n$ go down rapidly outside the sparse set.

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