# How slowly can a series grow to be convergent?

This may be a poorly worded question, but I hope to flesh out my ideas well. The takeaway here is this: some series diverge to infinity while others converge to a fixed value. Take the two classical examples: the harmonic series and the Basel Problem

$$\displaystyle\sum_{n=1}^\infty \frac{1}{n}= \infty \, , \hspace{0.6cm} \displaystyle\sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6}$$

On one end, we have a divergent series, whereas on the other end, we have a convergent series, both series of which seem eerily similar, except for the square in the latter. A question I might raise would be: at what "rate of growth" (loosely speaking) does a series have to grow to tip over from the point of convergence to sudden, chaotic divergence?

Yes, in Calculus, you learn about various convergence tests that allow one to test whether a given series is convergent, but I am wondering if there is a famous "rate of growth" that a series must "exceed" in order to indisputably diverge.

• You raise a classical (1900s) question. And an instructive one at that Sep 4, 2021 at 3:14
• Consider $\sum\dfrac1{n\log n}$ and $\sum\dfrac1{n(\log n)^2}$, among others. Sep 4, 2021 at 3:21
• @A rural reader: You raise a classical (1900s) question. --- This should be 1800s. See my 10 September 1999 sci.math post on the topic. There are also several relevant MSE questions/answers, but I don't have time now to look them up. (A google search for "du Bois Reymond" and "convergence" and my name, which I knew would bring up one or more items with references, led me to the sci.math post in a few seconds.) Sep 4, 2021 at 3:48
• For explicit illustrative purposes, Consider that $\int_1^n (1/x) dx = \log(n)$. Also consider that between $1$ and $n$, you can construct $(n-1)$ rectangles, each of width $1$, each of whose right edge is on the point $(x = r ~: ~r \in \{2,3,\cdots,n\})$. Assuming that rectangle $r$ has a height of $(1/[r+1])$, you have that the collection of rectangles is completely under the curve $y = (1/x).$ This implies that $\log(n) > [(1/2) + (1/3) + \cdots + (1/n)].$ This implies (for example) that $(1 + 13.82) > [1 + \log (10^6)] > \sum_{k=1}^{10^6} (1/k).$ Sep 4, 2021 at 4:21
• Does this answer your question? Which series converges the most slowly? Sep 4, 2021 at 13:14

However, for practical purposes, it can be helpful to note that the series $$\sum_{n=2}^\infty \frac1{n\log n},\; \sum_{n=3}^\infty \frac1{n\log n\log\log n},\; \sum_{n=16}^\infty \frac1{n\log n\log\log n\log\log\log n},\; \dots$$ all diverge, while the series $$\sum_{n=2}^\infty \frac1{n(\log n)^{1+\varepsilon}},\; \sum_{n=3}^\infty \frac1{n\log n(\log\log n)^{1+\varepsilon}},\; \sum_{n=16}^\infty \frac1{n\log n\log\log n(\log\log\log n)^{1+\varepsilon}},\; \dots$$ all converge for every $$\varepsilon>0$$. (You can verify these assertions all using the Integral Test!)