Can a sequence which decays more slowly still yield a converging series?

Question

In Bergman's companion notes to Rudin, he says that "If a sequence of positive terms has convergent sum, so does every sequence of positive terms which decays more rapidly." So given a sequence $\{a_n\}$ of positive terms such that $\sum_n a_n$ converges, if $\{b_n\}$ is such that $$ \lim_{n\to\infty} \frac{a_n}{b_n} = +\infty, $$ then $\sum_nb_n$ converges. I can prove this given $\{a_n\}$ and $\{b_n\}$. However, can we find $b_n$ which decays more slowly, i.e. $$ \lim_{n\to\infty} \frac{b_n}{a_n} = +\infty $$ such that $\sum_n b_n$ converges?

Similarly, we have the claim "if a sequence of positive terms has divergent sum, then so does every sequence of positive terms which decays more slowly."

Answer 1

The question is the following:

Show that for every converging series $\sum\limits_na_n$ with positive terms there exists a converging series $\sum\limits_nb_n$ with positive terms such that $\lim\limits_{n\to\infty}\frac{a_n}{b_n}=0$.

A hands-on approach is as follows: for every $n$, consider $$A_n=\sum_{k\geqslant n}a_k,$$ then, by hypothesis, $A_n\to0$ hence, for each $k\geqslant0$, there exists some finite $\nu(k)$ such that $$A_{\nu(k)}\leqslant2^{-k}.$$ Assume without loss of generality that the sequence $(\nu(k))$ is nondecreasing and define $(b_n)$ by $$\color{red}{\forall k\geqslant0,\quad\forall n\in[\nu(k),\nu(k+1))},\quad \color{red}{b_n=k\,a_n}.$$ Then $\frac{a_n}{b_n}\to0$ when $n\to\infty$ and $$\sum_{n\geqslant\nu(0)}b_n=\sum_{k\geqslant0}k\sum_{n=\nu(k)}^{\nu(k+1)-1}a_n\leqslant\sum_{k\geqslant0}kA_{\nu(k)}\leqslant\sum_{k\geqslant0}k2^{-k},$$ which is finite, hence the series $\sum\limits_nb_n$ converges, as desired.

Answer 2

If I can remember, Rudin clearly shows that there is no threshold for convergence or divergence of positive series, in chapter 3 of his book. You can construct convergent series with arbitrary slow decay by multiplying by logarithms. For instance $$ \sum_n \frac{1}{n \log n \log \log n} $$ diverges, whereas $$ \sum_n \frac{1}{n \log n (\log \log n)^2} $$ converges, and so on.

Answer 3

You can build this using: $R_n = \sum_{k=n}^{+\infty} a_k \geq 0$ (exists since $\sum a_k$ converges)

$\sum a_n$ converges, i.e : $R_n \rightarrow 0$ , when $n\rightarrow +\infty$

You can define: $$b_n = \sqrt{R_n} -\sqrt{R_{n+1}}$$

Note that whenever $a_i = 0$ : $b_i = \sqrt{R_i} -\sqrt{R_{i+1}} =\sqrt{R_{i+1}} -\sqrt{R_{i+1}} = 0 $

So if you define: $v_n = \frac{a_n}{b_n} $ you can set: $v_i = 0$ when $a_i = 0$

Now when $a_n \neq 0$: $$v_n = \frac{a_n}{b_n} = \frac{R_n - R_{n+1}}{\sqrt{R_n} -\sqrt{R_{n+1}}} = \sqrt{R_n} +\sqrt{R_{n+1}} \rightarrow 0 , n\rightarrow +\infty $$

And $\sum b_n$ converges by a telescoping.

---

Edit: Likewise, you can also find a sequence ($c_n$) such as :$\frac{c_n}{a_n} \rightarrow 0$ and $\sum c_n$ converges.

You can consider : $c_n = R_n^2-R_{n+1}^2$ with the same statements as for $b_n$ when ($a_n$) contains zero terms:

$$ \frac{c_n}{a_n} = \frac{R_n^2-R_{n+1}^2}{a_n} = \frac{R_n^2-R_{n+1}^2}{R_n - R_{n+1}} = R_n + R_{n+1} \rightarrow 0 $$

Here again $\sum c_n$ converges by telescoping. So one cannot define the notion of fastest converging series, since given any converging one you can build a new that converges faster.

Answer 4

There are a couple of things... slightly off... in your understanding.

The theorem does not say, as you imply, that if $$\sum_{n=1}^\infty a_n$$ exists and if $$\lim_{n\to\infty}\frac{b_n}{a_n} = \infty,$$ then the sum $$\sum_{n=1}^\infty b_n$$ exists. I advise you to re-check what your notes say.