# Variant of Baby Rudin Ch 3 Ex 11

Suppose that $a_n > 0$. Then a part of a problem in Rudin asks if $\sum_n a_n = \infty$ implies that $\sum_n \frac{a_n}{1+na_n} = \infty$.

This question is nicely answered here: Convergence/divergence of $\sum\frac{a_n}{1+na_n}$ when $a_n\geq0$ and $\sum a_n$ diverges

But this solution is non-monotonic. My question is if there exists a monotonic sequence that makes this converge. I was trying to bound this below by something, to no avail. I then tried to subtract the summand by some $\frac{c_n}{n}$, where $c_n$ is bounded, but I was not able to get a difference that converged to imply divergence since $\sum_n \frac{c_n}{n} = \infty$. (I was trying $0 < c_n < M$, but I suppose that I should have tried $c_n$ bounded below since I want divergence of our modified harmonic series.) Obviously, such a method is doomed to fail since I am trying to prove that this always converges when obvious tries like $a_n = \frac{1}{n}$ makes our series diverge, so I have to creatively apply some sort of condition to make it converge (like $na_n \to 0$, which also doesn't work because of logs), but that did not work.

So, I was wondering if there was any ideas that anyone had. Thanks. :)

If $(a_n)$ is monotonic, then the divergence of $\sum a_n$ implies the divergence of $\sum \frac{a_n}{1 + n a_n}$.

First consider the case that $(a_n)$ is monotonically decreasing. Let

$$S_k = \sum_{n = 1}^k a_n.$$

By monotonicity, we have $n a_n \leqslant S_n$, and for large enough $n$ we have $S_n \geqslant 1$, so $1 + a_n \leqslant 2S_n$. Then

$$\sum_{n = m+1}^k \frac{a_n}{1 + n a_n} \geqslant \sum_{n = m+1}^k \frac{a_n}{2S_n} \geqslant \frac{1}{2S_k}\sum_{n = m+1}^k a_n = \frac{S_k - S_m}{2S_k} = \frac{1}{2} - \frac{S_m}{2S_k}$$

for all large enough $m$ and $k > m$.

Letting $k$ become really large, since $S_k \to +\infty$, we see that for all $m$ there is a $k > m$ with

$$\sum_{n = m+1}^k \frac{a_n}{1 + n a_n} > \frac{1}{3},$$

so

$$\tilde{S}_k := \sum_{n = 1}^k \frac{a_n}{1 + n a_n}$$

is not a Cauchy sequence.

If $(a_n)$ is monotonically increasing, then

$$\frac{a_n}{1 + n a_n} \geqslant \frac{a_n}{n(1 + a_n)} \geqslant \frac{a_1}{1 + a_1}\cdot \frac{1}{n}$$

(since $t \mapsto \frac{t}{1+t}$ is increasing) shows the divergence.

• I think the quantifier with $m$ and $k$ are a bit tricky and could perhaps be better explained, but it all works out. Very, very good answer. +1 Jun 13, 2017 at 20:10
• This assumes $a_n$ is monotonically decreasing. What if $a_n$ is monotonically increasing? Jun 13, 2017 at 20:10
• @MichaelLee Then $\frac{a_n}{1 + n a_n} \geqslant \frac{c}{n}$, so the series diverges too. Jun 13, 2017 at 20:13
• Yes, this works. Using monotonicity you get $S_n \geq na_n$ and basically use the fact that $\sum a_n / s_n$ diverges (which is part of the problem from Rudin). Thanks. :) Jun 13, 2017 at 20:18