3

Sorry if this is too simple of a question.

Let $S_n = \sum \limits _{k=0} ^{n} \dfrac{1}{k}$. For another sequence $\{x_k\}$ define $S'_n = \sum \limits _{k=0} ^{n} x_k$. Is it possible that $S'_n$ diverges and $ lim _{n \rightarrow \infty} \dfrac{S'_n}{S_n} = 0$?

  • 5
    $$\sum_{k=1}^{N}\frac{1}{k\log(k+1)}\sim \log\log N, $$ $$\sum_{k=1}^{N}\frac{1}{k\log(k+1)\log\log(k+1)}\sim\log\log\log N,$$ $$\sum_{k=1}^{N}\frac{\mu(k)}{\sqrt{k}}\sim\sqrt{\log\log N}$$ etcetera. There is no such thing as the slowest divergent series. – Jack D'Aurizio Apr 06 '18 at 11:29
  • Makes sense, thanks. –  Apr 06 '18 at 11:30
  • The reciprocals of the primes are pretty slow too... – abiessu Apr 06 '18 at 11:31
  • So then would it be possible for there to be a series that diverges more slowly than log log log ..... log N for any number of logs? –  Apr 06 '18 at 11:33
  • Well, for any particular number of logs, e.g. a million, you could add one more to get even slower divergence. It's rather like trying to find the largest integer, you can't because you can always add one. – badjohn Apr 06 '18 at 11:39
  • My idea is similar to the ordinals. It's true that you can always get a higher integer by adding 1, but starting with an integer and adding 1 will always give something smaller than $\omega$. Is it possible that there's a divergent series so that taking N and adding log a finite number of times will always give something that grows faster than that series? –  Apr 06 '18 at 11:47
  • Sorry, I cannot understand what you are asking for here. Several people have said that there is no slowest divergent series. I don't see that we can add much to that. A problem with iterated logs is the range on which they are defined. $\log(x)$ itself is only defined for $x > 0$ so $\log(\log(x))$ is only defined for $x > 1$, $\log(\log(\log(x)))$ is only defined for $x > 10$, etc. So, $\log^\omega$ won't be defined at all. – badjohn Apr 06 '18 at 14:53
  • @badjohn: on the other hand, for any $N>1$ you may define $\log^(N)$ as the number of logarithms needed to make $\log\log\ldots\log(N)$ less than $1$. Still we have a sequence ${a_n}_{n\geq 1}$ such that $A_N=o\left(\log^ N\right)$ by the construction below. – Jack D'Aurizio Apr 06 '18 at 15:26
  • @JackD'Aurizio Yes, I was just trying to explore what the OP wanted here. – badjohn Apr 06 '18 at 15:31
  • I guess it might be clearer to ask the negative of my question. Is it correct to say that for any divergent series, there exists a natural number $k$ such that $log^k N$ diverges more slowly than said series?

    Also, I really appreciate everyone's help and guidance.

     –  Apr 07 '18 at 08:11

2 Answers2

2

There actually is an interesting exercise in Baby Rudin asking to prove that if $\{a_n\}_{n\geq 1}$ is a (decreasing) sequence of positive real numbers such that $\lim_{n\to +\infty}a_n = 0$ and $\sum_{n\geq 1}a_n = +\infty$, there exists a (decreasing) sequence $\{b_n\}_{n\geq 1}$ of positive real numbers such that $\lim_{n\to +\infty} b_n = 0$, $\sum_{n\geq 1}b_n=+\infty$ and $\lim_{n\to +\infty}\frac{b_n}{a_n}=0$. Here it is a sketch of the proof:

["Deceleration" of divergent series] Let $A_n=a_1+a_2+\ldots+a_n$. $\{A_n\}_{n\geq 1}$ is an increasing sequence of positive numbers such that $A_n\to +\infty$ as $n\to +\infty$. By summation by parts

$$\begin{eqnarray*} \sum_{n=1}^{N}\frac{a_n}{\sqrt{A_n}}&=&\sqrt{A_N}-\sum_{n=1}^{N-1}A_n\left(\frac{\sqrt{A_n}-\sqrt{A_{n+1}}}{\sqrt{A_n A_{n+1}}}\right)\\&=&\sqrt{A_N}+\sum_{n=1}^{N-1}A_n\left(\frac{a_{n+1}}{\sqrt{A_n A_{n+1}}(\sqrt{A_n}+\sqrt{A_{n+1}})}\right)\\&\geq&\sqrt{A_N}+\sum_{n=1}^{N-1}\sqrt{\frac{A_n}{A_{n+1}}}\cdot\frac{a_{n+1}}{2\sqrt{A_{n+1}}}\end{eqnarray*} $$ hence $\sum_{n=1}^{N}\frac{a_n}{\sqrt{A_n}}\sim 2\sqrt{A_N}$ by Cesàro-Stolz and by taking $b_n=\frac{a_n}{\sqrt{A_n}}$ we are done.

More simply, you may consider a sequence of positive numbers $\{c_n\}_{n\geq 1}$ which is divergent to $+\infty$ as slowly as you like (for instance $c_n=\log\log\ldots\log(n+K)$), then define $a_n$ as $c_{n+1}-c_n$. By telescoping $A_n$ diverges exactly like $c_n$.

Jack D'Aurizio
  • 353,855
0

The concrete examples are very interesting. Still: if $S_n$ is increasing to $\infty$, can we find $S'_n$ increasing to $\infty$ with $\frac{S'_n}{S_n}\to 0$? Of course. You can take $S_n'= \sqrt{S_n}$, or $\log S_n$, etc.

orangeskid
  • 53,909