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

Solution 1:

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.

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

$$\rm \int_{a}^\infty \frac{dx}{x \log x\ \log_2 x\ \cdots\ \log_{k-1} x\ (\log_k x)^s } $$

Solution 2:

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.

Solution 3:

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).

Solution 4:

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.

Solution 5:

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 $$ \sum_{n=2}^\infty \frac{1}{n(\log n)^s}\quad\Longleftrightarrow\quad \sum_{k=1}^\infty \frac{1}{k^s}\quad\Longleftrightarrow\quad s>1. $$