Is there a sequence $a_n$ decreasing to $1$ such that $\sum n^{-a_n}$ converges?

Question

Does there exist a decreasing sequence $\{a_n\}$ such that $\lim_{n\to\infty}a_n = 1$ for which the series $$\sum_{n=1}^{\infty} \frac{1}{n^{a_n}}$$ converges? Inspiration: I was a bit surprised to learn that this series diverges for $a_n = 1+\frac{1}{n}$. Now I'm curious if anything strange happens here, if there is a sequence $\{a_n\}$ that decreases to $1$ "slowly enough" to make the series converge. Otherwise we get a slight generalization of what we know about $p$-series from this.

Answer 1

Let $$a_n = 1 + \frac{2\ln\ln n}{\ln n}\text{ for }n\geq 3.$$ Then $\lim_{n\to\infty}a_n = 1$. To see that the series converges, note that for $n\geq 3$, we have $$\frac{1}{n^{a_n}} = \frac{1}{n^{1+\frac{2\ln\ln n}{\ln n}}} = \frac{1}{n\cdot e^{2\ln\ln n}} = \frac{1}{n(\ln n)^2}.$$ Thus, we have $$\sum_{n=3}^{\infty} \frac{1}{n^{a_n}} = \sum_{n=3}^{\infty} \frac{1}{n(\ln n)^2}.$$ Consider this improper integral $$\int_{3}^{+\infty} \frac{\mathrm{d} x}{x(\ln x)^2}=\int_3^{+\infty} \frac{\mathrm{d}(\ln x)}{(\ln x)^{2}}=\int_{\ln 3}^{+\infty} \frac{\mathrm{d} u}{u^{2}}<+\infty.$$ It follows that $$\sum_{n=3}^{\infty} \frac{1}{n(\ln n)^2}<+\infty.$$

Answer 2

For $2^{k}<n\le 2^{k+1}$ let $a_n=1+k^{-1/2}.$ Then $$\sum_{n=2^{k}+1}^{2^{k+1}}{1\over n^{a_n}}\le {2^k\over 2^{k(1+k^{-1/2})}}={1\over 2^{k^{1/2}}}$$

Answer 3

General observation based on this answer, if you let $$a_n = 1 + \frac{\ln(b_n)}{\ln(n)}$$ for some sequence $\{b_n\}$, then this series can be re-writtten as $$\sum \frac{1}{n\,b_n}\,,$$ the convergence/divergence of which is easier to think about.