# Is there a slowest rate of divergence of a series?

$$f(n)=\sum_{i=1}^n\frac{1}{i}$$ diverges slower than $$g(n)=\sum_{i=1}^n\frac{1}{\sqrt{i}}$$ , by which I mean $\lim_{n\rightarrow \infty}(g(n)-f(n))=\infty$. Similarly, $\ln(n)$ diverges as fast as $f(n)$, as $\lim_{n \rightarrow \infty}(f(n)-\ln(n))=\gamma$, so they 'diverge at the same speed'.

I think there are an infinite number of 'speeds of divergence' (for example, $\sum_{i=1}^n\frac{1}{i^k}$ diverge at different rates for different $k<1$). However, is there a slowest speed of divergence?

That is, does there exist a divergent series, $s(n)$, such that for any other divergent series $S(n)$, the limit $\lim_{n \rightarrow \infty}(S(n)-s(n))=\infty$ or $=k$? If so, are there an infinite number of these slowest series?

• The answer is no. If you would like to see this yourself, given any arbitrary divergent series, try to come up with another divergent series that diverges more slowly. – Lord Soth Jul 25 '13 at 16:46
• @LordSoth I thought so, but how do I know for sure there will always be a slower one? – Meow Jul 25 '13 at 16:47
• See, for example, the third paragraph of this question: mathoverflow.net/questions/49415/… – Lord Soth Jul 25 '13 at 16:49
• I've added the tag "asymptotics" because this seems to be what you are getting at - the order of growth of such series, and related questions is a huge subject, and there is a whole set of notation ("Big O" and "little o") around such comparisons. Someone may post some better links to basic material. But note, for example, that $f(n)$ grows like $\ln (n)$. – Mark Bennet Jul 25 '13 at 16:52
• There you go, I was searching for you the following nice paper: math.depaul.edu/mash/worstconv.pdf – Lord Soth Jul 25 '13 at 16:54

The proof in the paper Neither a worst convergent series nor a best divergent series exists" by J. Marshall Ash that I referenced above is so nice that I wanted to reproduce it below before it gets lost on the Internet.

$\bf{Theorem: }$ 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 $\lim_{n\rightarrow\infty} C_n/c_n = \infty$. Similarly, for any divergent series $\sum_{n=1}^{\infty} D_n$ with positive terms, there exists a divergent series $\sum_{n=2}^{\infty} d_n$ with much smaller terms in the sense that $\lim_{n\rightarrow\infty} \frac{d_n}{D_n} = 0$.

$\bf{Proof: }$ For each $n$, let $r_n = c_n + c_{n+1}+\cdots$ and $s_n = D_1 + \cdots + D_n$. Letting $C_n = \frac{c_n}{\sqrt{r_n}}$ and $d_n = \frac{D_n}{s_{n-1}}$, then $\lim_{n\rightarrow\infty} \frac{C_n}{c_n} = \lim_{n\rightarrow\infty} \frac{1}{\sqrt{r_n}}=\infty$ and $\lim_{n\rightarrow\infty} \frac{d_n}{D_n} = \lim_{n\rightarrow\infty} \frac{1}{s_{n-1}} = 0$, so it only remains to check $\sum C_n$ converges and that $\sum d_n$ diverges. To see that this is indeed the case, simply write $C_n = (1/\sqrt{r_n})(r_n-r_{n+1})$ and $d_n = 1/s_{n-1}(s_n-s_{n-1})$; observe that $\int_0^{r_1} 1/\sqrt{x}dx<\infty$ and $\int_{s_1}^{\infty} 1/xdx = \infty$; and note that the $n$th term of series $\sum C_n$ is the area of the gray rectangle in Figure 1a, while the $n$th term of series $\sum d_n$ is the area of the gray rectangle in Figure 1b.

• Ash's paper is on JSTOR reference being The College Mathematics Journal, Vol. 28, No. 4 (Sep., 1997), pp. 296-297 with the stable URL being jstor.org/stable/2687153 – Fixed Point Jul 26 '13 at 21:31

Just to expand on Raymond Manzoni and to partially answer your question with one of the coolest things I ever learned in my early childhood (as a math undergrad),

consider the family of series,

$$\begin{eqnarray} \sum \frac{1}{n}&=&\infty \\ \sum \frac{1}{n\ln(n)}&=&\infty \\ \sum \frac{1}{n\ln(n)\ln(\ln(n))}&=&\infty \end{eqnarray}$$

and so on. They all diverge and you can use the integral test to easily prove this. But here is the kicker. The third series actually requires a googolplex numbers of terms before the partial terms exceed 10. It is only natural that if natural log is a slowly increasing function, then log of log diverges even slower.

On the other hand, consider the family,

$$\begin{eqnarray} \sum \frac{1}{n^2}&<&\infty \\ \sum \frac{1}{n(\ln(n))^2}&<&\infty \\ \sum \frac{1}{n\ln(n)(\ln(\ln(n)))^2}=38.43...&<&\infty \end{eqnarray}$$

all converge which can be easily verified easily again using the integral test. But the third series converges so slowly that it requires $10^{3.14\times10^{86}}$ terms before obtaining two digit accuracy presented. Talking about getting closer to the "boundary" between convergence and divergence. Using this you can easily make up a series to converge or to diverge as slow as you want. So to answer your question, no there is no such thing as "the slowest diverging series". Any slowly diverging series you pick, we can come up with one diverging even slower.

Reference: Zwillinger, D. (Ed.). CRC Standard Mathematical Tables and Formulae, 30th ed. Boca Raton, FL: CRC Press, 1996.

• "So to answer your question, no there is no such thing..." That is shown in answers of Clement C. and Lord Soth. The examples in this answer are interesting, but I do not see how they exactly support the assertion. – Jonas Meyer Jul 26 '13 at 4:17
• I am just giving a set of examples which include series converging and diverging arbitrarily slowly. This isn't really a proof, like I said, just examples. – Fixed Point Jul 26 '13 at 4:19
• "arbitrarily slowly": I see that each converges more slowly than the last, but I don't know what "arbitrarily slowly" means in the context of the question. Does it mean that given an arbitrary divergent series $\sum\limits_{n=1}^\infty a_n$ of positive numbers, there exists a series in your set of examples that diverges more slowly? – Jonas Meyer Jul 26 '13 at 4:22

The Cauchy condensation test allows to find an infinity of slower divergent series :

• $\displaystyle \sum \frac 1n\;$ behaves like $\;\displaystyle \sum 2^n \frac 1{2^n}=\sum 1\;$ which diverges
• $\displaystyle \sum \frac 1{n\ln(n)}\;$ behaves like $\;\displaystyle \sum 2^n \frac 1{2^n\;n}=\sum \frac 1n\;$ which diverges
• $\displaystyle \sum \frac 1{n\ln(n)\ln(\ln(n))}\;$ behaves like $\;\displaystyle \sum 2^n \frac 1{2^n\;n\ln(n)}=\sum \frac 1{n\ln(n)}\;$ which diverges... QED.

(from Chaitin's book 'Algorithmic Information Theory')

• Interesting example of something, but not an answer to the question. – Jonas Meyer Jul 25 '13 at 17:37
• @JonasMeyer: Well yes it shows just a rather neat family at the 'border of divergence' (and there is a corresponding family on the other side...). – Raymond Manzoni Jul 25 '13 at 18:17
• RaymondManzoni: I agree it is neat. What exactly is meant by 'border of divergence'? – Jonas Meyer Jul 26 '13 at 4:19
• @Jonas: It was just a picture. Formalizing this would ask for much more work and start with something like "Consider the space of elementary functions of $n$ from the point of view of differential algebra that is the logarithmic, exponential and algebraic extensions of (say) the field of rational functions over $\mathbb{Q}$ (still not considering all the series and omitting the even/odd $n$ sequences and so on). Each of these series should have a 'speed' (to be defined...) of convergence or divergence with $0$ between the two sets. – Raymond Manzoni Jul 26 '13 at 9:57
• From this simple point of view the different series proposed will approach $0$ from the negative side and the replacement of any 'exterior' term $n, \ln(n), \ln(\cdots\ln(n)\cdots)$ by the same term to a power larger than $1$ bring it on the positive convergent side. But I can't be satisfied with this $1D$ picture nor with the infinite (countable) dimensional view of all the possibles elementary functions of $n$ ; at least a second dimension with the value of the limit would make the 'thing' nicer but oblige me to define a finite limit for all divergent series. Fine I'll have to read Hardy ! – Raymond Manzoni Jul 26 '13 at 10:03

If you have $(a_n)_{n\in\mathbb{N}}$ a divergent series: $$\sum_{n=0}^N a_n \xrightarrow[N\to\infty]{}\infty$$ with $a_n > 0$ for all $n\in\mathbb{N}$), consider the series with term $$b_n\stackrel{\rm{}def}{=}\frac{a_n}{\sum_{k=0}^n a_k}.$$ One can show that the series $\sum b_n$ diverges.

• I don't know how you could begin showing that (I know very little about asymptotics). Could you hint or provide a link? – Meow Jul 25 '13 at 16:58
• Mmmh... there should be a lot of answers to this on the site, but essentially, you can first see that is $b_n\not\to 0$, it's settled; so assume $b_n\to 0$. Then $b_n\sim -\ln(1-b_n)$, and writing further: $$-\ln(1-b_n) = -\ln(1-\frac{a_n}{A_n}) = -\ln\frac{A_{n-1}}{A_n} = \ln A_n - \ln{A_{n-1}}$$ where $A_n=\sum_{k=0}^n a_k$. Now, since $\sum a_n$ diverges, $\ln A_n\to\infty$, and by telescoping the terms this means that the series $\sum(-\ln(1-b_n)) = \sum ( \ln A_n - \ln A_{n-1} )$ diverges; therefore, since $b_n\sim -\ln(1-b_n)$, the series $\sum b_n$ diverges as well. – Clement C. Jul 25 '13 at 17:00

Suppose there is a slowest-diverging series. Call it $(a_n)$.

Define $A(n)=\sum a_n$. Thus, $A(n)$ is a slowly-diverging function.

Define $B(n)=\ln A(n)$. Thus, $B(n)$ diverges even slower than $A(n)$.

Define $b_n=B(n)-B(n-1)$. Thus, $(b_n)$ is an even slower diverging series. Contradiction.

Let A(n) be an Ackermann function, let $1_{A}$ be its indicator function (i.e $1_{A} (x)$=1 if A(n)=x for some n ; 0 otherwise). Then $\sum1_{A}^{}$ diverges. A one that diverges faster but more interesting is $s(0)=1, s(n+1)= s(n)+ \frac{1}{p_{n}s(n)}$ where $p_{n}$ is the nth prime.

• How does this address the question given by the OP? – Christopher K Apr 3 '14 at 0:26
• Thinking outside the box. One is the slowest concrete example posted. It also shows the futility of such search. The other also is slow and uses some neat math. – user139964 Apr 3 '14 at 0:58
• Don't presume that I gave you the downvote. I didn't. – Christopher K Apr 3 '14 at 1:00
• I didn't; thanks for the feedback. – user139964 Apr 3 '14 at 1:19
• Upvote (+1)! The answer makes sense. I'd never heard of the Ackermann function. Where does this come up? – Christopher K Apr 3 '14 at 1:30