Looking at slow divergent series.

Question

So today i have two questions in one, basically i need explanations. It is school break and where can i find a better place to tutor myself with math apart from here. Now I came across this topic of divergent series,

I was wondering apart from; $1+\frac{1}{2}+\frac{1}{3}...$ what other slowly divergent series do we have?

Thanks.

Answer 1

An interesting family of diverging series is as follows: $$ \sum_{n=1}^\infty \frac 1{n}\\ \sum_{n=1}^\infty \frac 1{n\times \ln(n)}\\ \sum_{n=1}^\infty \frac 1{n\times \ln(n)\times \ln(\ln(n))}\\ \sum_{n=1}^\infty \frac 1{n\times\ln(n)\times\cdots\times\ln(\ln(\cdots\ln(n)\cdots))} $$ In order to prove that these diverge, it helps to apply the Cauchy condensation test.

Answer 2

Let's define the iterated logarithm by $\ln_1 x = \ln x$ and for $p \ge 1$ integer $\ln_{p+1} x = \ln(\ln_p x)$ which is defined for $x$ large enough.

One can prove that $$\sum \frac{1}{n \ln n \ln_2 n \dots \ln_p n}$$ diverges slower than $$\sum \frac{1}{n \ln n \ln_2 n \dots \ln_{p-1} n}$$ that diverges slower than ... that diverges slower than $$\sum \frac{1}{n}$$

Answer 3

If you read the other answers involving divergent series with logarithm and want a series that diverges even slower, consider defining $\sum_n 1/f(n)$ where $f(x)$ is defined as $1$ if $x \leq e$, and otherwise defined as $f(x) = x f(\log x)$. In otherwords, as $n$ grows, you keep taking the log over and over again in the denominator and multiplying as long as the rules allow. It is challenging (a high-end Putnam math contest problem) to show this series diverges, and it turns out that it diverges asymptotically like $\log^* (n)$, where $\log^*(n)$ is the number of times you need to take the logarithm iteratively starting with $n$ in order to get an answer $\leq 1$. A very, very slow growing function indeed because the logarithms of large numbers are so much smaller.