Is there a slowest rate of divergence of a series?

diverges slower than
, by which I mean lim. Similarly, \ln(n) diverges as fast as f(n), as \lim_{n \rightarrow \infty}(f(n)-\ln(n))=\gamma, so they ‘diverge at the same speed’.

I think there are an infinite number of ‘speeds of divergence’ (for example, \sum_{i=1}^n\frac{1}{i^k} diverge at different rates for different k<1). However, is there a slowest speed of divergence?

That is, does there exist a divergent series, s(n), such that for any other divergent series S(n), the limit \lim_{n \rightarrow \infty}(S(n)-s(n))=\infty or =k? If so, are there an infinite number of these slowest series?


The proof in the paper ``Neither a worst convergent series nor a best divergent series exists" by J. Marshall Ash that I referenced above is so nice that I wanted to reproduce it below before it gets lost on the Internet.

\bf{Theorem: } Let \sum_{n=1}^{\infty} c_n be any convergent series with positive terms. Then, there exists a convergent series \sum_{n=1}^{\infty} C_n with much bigger terms in the sense that \lim_{n\rightarrow\infty} C_n/c_n = \infty. Similarly, for any divergent series \sum_{n=1}^{\infty} D_n with positive terms, there exists a divergent series \sum_{n=2}^{\infty} d_n with much smaller terms in the sense that \lim_{n\rightarrow\infty} \frac{d_n}{D_n} = 0.

\bf{Proof: } For each n, let r_n = c_n + c_{n+1}+\cdots and s_n = D_1 + \cdots + D_n. Letting C_n = \frac{c_n}{\sqrt{r_n}} and d_n = \frac{D_n}{s_{n-1}}, then \lim_{n\rightarrow\infty} \frac{C_n}{c_n} = \lim_{n\rightarrow\infty} \frac{1}{\sqrt{r_n}}=\infty and \lim_{n\rightarrow\infty} \frac{d_n}{D_n} = \lim_{n\rightarrow\infty} \frac{1}{s_{n-1}} = 0, so it only remains to check \sum C_n converges and that \sum d_n diverges. To see that this is indeed the case, simply write C_n = (1/\sqrt{r_n})(r_n-r_{n+1}) and d_n = 1/s_{n-1}(s_n-s_{n-1}); observe that \int_0^{r_1} 1/\sqrt{x}dx<\infty and \int_{s_1}^{\infty} 1/xdx = \infty; and note that the nth term of series \sum C_n is the area of the gray rectangle in Figure 1a, while the nth term of series \sum d_n is the area of the gray rectangle in Figure 1b.

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Source : Link , Question Author : Meow , Answer Author : Lord Soth

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