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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.

Ben Grossmann
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    I suggest that you break your question into two different questions. The second part (with $\sqrt{-1}$) is not really a question about divergent series. – Michael Burr Jun 30 '15 at 20:46
  • It would be good to define precisely what you mean by "slowly divergent." Of course, there are infinitely many series which diverge more slowly than $1+\frac{1}{2}+\frac{1}{3}+\dots$. Consider, for example, $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\dots$ or indeed the entire class $\frac{1}{n}\left(1+\frac{1}{2}+\frac{1}{3}+\dots\right)$ where $n>1$ (and we know that these are divergent because multiplying each term in a divergent series by the same constant yields another divergent series). These all still grow logarithmically; you can get even slower divergent growth rates. – hexaflexagonal Jun 30 '15 at 20:50
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    To your second question: see this post – Ben Grossmann Jun 30 '15 at 20:51
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    "So today i have two questions in one". Avoid doing so in the future. – Ben Grossmann Jun 30 '15 at 20:58
  • Okay thanks. @Omnomnomnom I didnt know the rules, i thought asking two questions is not allowed. Thanks for the link i will read about $i$ –  Jun 30 '15 at 21:00
  • @Liam feel free to post your second question separately, though – Ben Grossmann Jun 30 '15 at 21:01
  • Okay thanks. Let me post. I want to read more on the same. –  Jun 30 '15 at 21:02
  • I've edited the second question out; this question has been posted separately over here – Ben Grossmann Jun 30 '15 at 21:17
  • Thanks very much. This site is a blessing . –  Jun 30 '15 at 21:20

3 Answers3

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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.

Ben Grossmann
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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.

user2566092
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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 $$\sum \frac{1}{n}$$

mathcounterexamples.net
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