# Convergence of the series $\sum \limits_{n=2}^{\infty} \frac{1}{n\log^s n}$

We all know that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s}$ converges for $s>1$ and diverges for $s \leq 1$ (Assume $s \in \mathbb{R}$).

I was curious to see till what extent I can push the denominator so that it will still diverges.

So I took $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\log n}$ and found that it still diverges. (This can be checked by using the well known test that if we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges).

No surprises here. I expected it to diverge since $\log n$ grows slowly than any power of $n$.

However, when I take $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n(\log n)^s}$, I find that it converges $\forall s>1$.

(By the same argument as previous).

This doesn't make sense to me though.

If this were to converge, then I should be able to find a $s_1 > 1$ such that

$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$

Doesn't this mean that in some sense $\log n$ grows faster than a power of $n$?

(or)

How should I make sense of (or) interpret this result?

(I am assuming that my convergence and divergence conclusions are right).

• They are right. You don't seem to have completed one of your sentences: "if this were to converge, then I should be able to find a s_1 > 1 such that"...? – Qiaochu Yuan Nov 11 '10 at 21:24
• There seems to be some bug with this. It doesn't display the entire thing. – user17762 Nov 11 '10 at 21:25
• @Sivaram: Yes even i tried editing it, and it still has not worked – anonymous Nov 11 '10 at 21:27
• There seems to be a bug/feature! when dealing with <. Probably confusion with html tags. Sivaram, please use \log n instead of just logn. – Aryabhata Nov 11 '10 at 21:32
• @ Chandru and Moron: Thanks for editing it. Now it looks fine. – user17762 Nov 11 '10 at 21:33

## 6 Answers

Yes $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n (\log n)^s}$ is convergent if $\displaystyle s > 1$, we can see that by comparing with the corresponding integral.

As to your other question, if

$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$

does not imply $\log n$ grows faster than a power of $\displaystyle n$. You cannot compare them term by term.

What happens is that the first "few" terms of the series dominate (the remainder goes to 0). For a small enough $\displaystyle \epsilon$, we have that $\log n > n^{\epsilon}$ for a sufficient number of (initial) terms, enough for the series without $\log n$ to dominate the other.

• Sorry. I meant $\exists s_1 > 1$ such that $\displaystyle \sum_{2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{2}^{\infty} \frac{1}{n(\log n)^{s}}$ – user17762 Nov 11 '10 at 21:56

This is a special case of classic general logarithmic convergence tests. Some early work in asymptotics was motivated by attempts to determine the "boundary of convergence" in terms of various functions, e.g. log-exp functions. Below is a a very interesting excerpt from Hardy's classic "A course in pure mathematics". For further work see his "Orders of infinity" and see especially this very interesting paper by G. Fisher on Paul du Bois-Reymond's work on the boundary between convergence and divergence (where he discovered diagonalization before Cantor).

Note in particular the following very general result that is presented in the excerpt: the following series and integral are convergent if $\rm\ s > 1\$ and divergent if $\rm\: s\le 1\$ where $\rm\: n_0,\ a\$ are any numbers large enough to ensure positivity of $\rm\ log_k x := \log \log \cdots \log x\:,\$ iterated $\rm\:k\:$ times.

$$\rm \sum_{n_0}^\infty \frac{1}{n \log n\ \log_2 n\ \cdots\ \log_{k-1} n\ (\log_k n)^s }$$

$$\rm \int_{a}^\infty \frac{dx}{x \log x\ \log_2 x\ \cdots\ \log_{k-1} x\ (\log_k x)^s }$$     • Off-topic question: What text is the excerpt from? Thank you! – Ayesha Jan 29 '14 at 1:04
• @Ayesha I say that in the text above it: Hardy's .... – Bill Dubuque Jan 29 '14 at 1:09
• Sorry, sorry, I somehow glanced over that. Thank you again for responding to my rather asinine question. – Ayesha Jan 29 '14 at 1:43

Du Bois-Reymond was interested in this question, positing some sort of series "between" all convergent series and all divergent series. Hausdorff didn't like that idea, and showed that given any countable sequences of convergence series converging "ever more slowly" and divergent series diverging "ever more slowly", you can find a convergent and a divergent series "in between". Here the order relation is the one given by the (conclusive) ratio test.

Bill's remark is related to the recursive Elias codes. The first few codes correspond to the convergent series

$\displaystyle \frac{1}{n(\log n)^2}, \frac{1}{n\log n (\log \log n)^2}, \ldots$

He then diagonalizes to obtain his $\omega$ code (Elias omega code, if you want to look it up), which corresponds to

$\displaystyle \frac{1}{n\log n\log \log n \cdots (\log^* n)^2},$

where $\log^* n$ is the number of times you need to apply $\log$ until you get below some constant (the other logs continue up to this constant); confusingly, coding theorists use $\log^* n$ to mean the $\log n \log\log n \cdots$ part. We can continue the iteration like this:

$\displaystyle \frac{1}{n\log n\log\log n \cdots \log^*n (\log \log^* n)^2}, \frac{1}{n\log n\log\log n \cdots \log^*n \log \log^* n (\log\log \log^* n)^2}, \cdots$

and so on. We can diagonalize again to obtain an $\omega + \omega$ code, and so on at least for $n \omega + m$. We an probably continue even to $\omega^2$ and beyond.

The corresponding divergent series are $\displaystyle \frac{1}{n\log n}, \frac{1}{n\log n\log\log n}, \ldots, \frac{1}{n\log n\log\log n \cdots \log^* n}, \frac{1}{n\log n\log\log n \cdots \log^* n \log\log^* n}, \ldots$

One can ask whether there is a scale for convergent series, that is a chain of (mutually comparable) series converging more and more slowly, such that for each convergent series there's one in the chain converging even more slowly; you can ask the same question about divergent series. Surprisingly, the existence of these is independent of ZFC (they exist given CH, and models in which they don't exist can be constructed using forcing).

• Indeed, it deserves to be better known that Du Bois Reymond - not Cantor - is the true discoverer of diagonalization - in this context, i.e. asymptotic growth rates of functions. – Bill Dubuque Nov 11 '10 at 22:53
• Thanks Bill and Yuval. I was looking for series which diverges slower and slower and your answer helps me on that note. – user17762 Nov 11 '10 at 23:09

Another test that applies to series of positive decreasing terms (and in this particular one in a rather elegant fashion) is the following: $$\sum_{n=1}^\infty a_n<\infty \quad\Longleftrightarrow\quad\sum_{k=1}^\infty 2^ka_{2^k}<\infty.$$ In our case $$\sum_{k=1}^\infty 2^ka_{2^k}=\sum_{k=1}^\infty 2^k\frac{1}{2^k(\log 2^k)^s}= \frac{1}{(\log 2)^s}\sum_{k=1}^\infty \frac{1}{k^s},$$ and thus $$\sum_{n=2}^\infty \frac{1}{n(\log n)^s}\quad\Longleftrightarrow\quad \sum_{k=1}^\infty \frac{1}{k^s}\quad\Longleftrightarrow\quad s>1.$$

• Monotonicity (as is in this case) is required? – Martín-Blas Pérez Pinilla Jan 31 '14 at 19:40
• @Martín-BlasPérezPinilla: I do not know the most general version of this criterion, but the version I know requires $a_n$ to be non-negative and decreasing. – Yiorgos S. Smyrlis Jan 31 '14 at 19:46
• Yes, now I remember I've seen in Baby Rudin. – Martín-Blas Pérez Pinilla Jan 31 '14 at 19:56

By the integral comparison test the given series is convergent if and only if the integral $$\int_2^\infty\frac{dx}{x\log^s(x)}$$ is convergent and notice that $\log'(x)=\frac1x$ so

• if $s>1$ $$\int_2^\infty\frac{dx}{x\log^s(x)}=\frac1{1-s}\log^{1-s}(x)\Big|_2^\infty<+\infty$$ so the series is convergent
• if $s=1$ $$\int_2^\infty\frac{dx}{x\log(x)}=\log(\log(x))\Big|_2^\infty=+\infty$$ so the series is divergent.
• The exponent $s$ is just on the log term, not the entire term – user2566092 Jan 31 '14 at 19:38
• Ah I'll edit my answer! – user63181 Jan 31 '14 at 19:41
• @user2566092 I edited my answer. – user63181 Jan 31 '14 at 19:50
• What about $s < 1$? – mavavilj Sep 19 '15 at 17:25

Convergence does not change if you switch to base-2 logs. There is a theorem that says that if $a_n$ is a decreasing positive sequence, then $\sum_n a_n$ converges if and only if $\sum_k 2^k a_{2^k}$ converges. If you apply this theorem's transformation to your sequence, you get $\sum_k 1/k^s$, which obviously converges if $s > 1$ and diverges if $s \leq 1$. If you'd like a proof of that theorem (called the Cauchy condensation test), I can provide it.