4

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?

Matt Rosenzweig
  • 4,201
  • 1
    See here http://math.stackexchange.com/questions/214556/if-a-n-is-not-summable-neither-is-left-fraca-n1-a-n/214570#214570 – blindman Nov 15 '12 at 03:46
  • 1
    http://math.stackexchange.com/questions/130471/convergence-divergence-of-sum-fraca-n1na-n-when-sum-a-n-diverges – Alti Nov 15 '12 at 03:48

2 Answers2

6

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

Robert Israel
  • 448,999
2

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.

André Nicolas
  • 507,029