$\sum_n a_n < \infty \quad \Leftrightarrow \quad \lim_{n\rightarrow \infty} \phi(n)a_n = \text{const.}$

convergence-divergenceintegrationsummation

I'm wondering how the following statement can be disproved.

Let $a_n$ be a positive sequence and assume there exists a function $\phi(n)$ with the properties
$$\sum_n a_n < \infty \quad \Leftrightarrow \quad \lim_{n\rightarrow \infty} \phi(n)a_n = \text{const.} \, .$$
I want to show, no such function can exist.

This question is along the lines of the answer given by Winther: If $\sum_{n_0}^{\infty} a_n$ diverges prove that $\sum_{n_0}^{\infty} \frac{a_n}{a_1+a_2+…+a_n} = +\infty $. He gives a proof for the case, where the constant is $0$. I'm wondering if the positive constant case makes a difference.

Just a few observations:

  1. $\phi(n)$ clearly must be asymptotically increasing, as $a_n$ is a null-sequence.
  2. The sequence $$a_n=\frac{{\rm d}}{{\rm d}x} \left(\log(\log(…\log(x)))\right)^s \Bigg|_{x=n} = \frac{{\rm d}}{{\rm d}x} \left(\log^{(m)}(x)\right)^s \Bigg|_{x=n} = \frac{s}{n\left(\log^{(m)}(n)\right)^{1-s}} \left(\prod_{k=1}^{m-1} \log^{(k)}(n) \right)^{-1}$$
    converges (by the integral criterion) for $s<0$ and diverges if $s>0$, where $m\in \mathbb{N}$. Because of this, if such a function existed, it would formally have the form $$\phi(n) \simeq n \prod_{k=1}^{\infty} \log^{(k)}(n) \, , \tag{1}$$
    not worrying about convergence of the product or if it is defined otherwise.
  3. The difference to Abel's proof is, that the sum $$\sum_n \frac{1}{\phi(n)}$$ would actually converge, because $\phi(n) \frac{1}{\phi(n)}=1$, not leading to the desired contradiction as in Abel's proof. Whether the sum, with $\phi(n)$ as in (1), would formally converge or not… I wouldn't know how to make sense of it.

Best Answer

(thanks to Martin R for pointing to an answer that gives a proof and a reference for the theorem used below)

The question you point to explains that this is related to the historical search for a function $\phi$ such that a series with positive terms $a_n$ is convergent iff $\lim_n\phi(n)a_n=0$. Both this case and your attempt to relax the condition to $\lim_n\phi(n)a_n=L<\infty$ can be dealt with thanks to the following theorem, proved for instance in [1]:

Theorem

Let $\sum_{n=1}^\infty c_n$ be any convergent series with positive terms. Then there exists a convergent series $\sum_{n=1}^\infty C_n$ with much bigger terms in the sense that

$$\lim_{n\to\infty}\frac{C_n}{c_n}=\infty$$

Similarly, for any divergent series $\sum_{n=1}^\infty D_n$ with positive terms, there exists a divergent series $\sum_{n=1}^\infty d_n$ with much smaller terms in the sens that

$$\lim_{n\to\infty}\frac{d_n}{D_n}=0$$

Assume that such a $\phi$ exists, with your relaxed condition. Then assuming there exists a convergent series for which $\lim_n \phi(n)a_n=L>0$, we can find a convergent series with terms $A_n$, such that $\frac{A_n}{a_n}\to\infty$. Therefore, $\lim_n \phi(n)A_n=\lim_n (\phi(n)a_n)\frac{A_n}{a_n}=\infty$, which is impossible since we are supposed to have $\lim_n \phi(n)A_n=L'<\infty$. So, the limit $L$ has to be zero, for all convergent series.

But even then, it won't work. First note that WLOG we can replace $\phi$ with $|\phi|$, so we can assume $\phi$ is positive. Consider the series with positive terms $D_n=\frac{1}{\phi(n)}$. Since $\phi(n)D_n\to1$, the series is divergent. Therefore, we can find another divergent series with positive terms $d_n$, such that $\frac{d_n}{D_n}\to0$. But then $\lim_n\phi(n)d_n=\lim_n(\phi(n)D_n)\frac{d_n}{D_n}=0$, which contradicts the fact that the series is divergent.

Therefore, such a $\phi$ can't exist.

Note that such a $\phi$ is akin to a fixed boundary between convergent and divergent series, and what the theoreom above say is such a boundary doesn't exist.

[1] J. Marshall Ash ,"Neither a Worst Convergent Series nor a Best Divergent Series Exists", The College Mathematics Journal Vol. 28, No. 4 (Sep., 1997), pp. 296-297, JSTOR 2687153, DOI 10.2307/2687153

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