I'm trying to come up with a positive sequence $\{a_n\}_1^\infty$ such that $\lim_{n\to\infty} \left(\sqrt[\leftroot{-2}\uproot{2}n]{a_n}\right) = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$, but $\forall \alpha > 0$ the series $\sum_1^\infty a_n n^\alpha$ converges. I know $1/n^n$ goes to zero faster than $n^\alpha$ goes to infinity, but it converges by the two tests. I've tried screwing around with $1/n^n$ but had no luck. Any thoughts?


1 Answer 1


If $a_n=e^{-\sqrt{n}}$ then $$ \lim_{n\to\infty}a_n^{\frac{1}{n}}=\lim_{n\to\infty}e^{-\frac{1}{\sqrt{n}}}=1$$ and $$ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\exp[\sqrt{n+1}-\sqrt{n}]=1$$ However, $$ \lim_{x\to\infty}x^{\beta}e^{-\sqrt{x}}=\lim_{y\to\infty}y^{2\beta}e^{-y}=0$$ for all $\beta>0$, so taking $\beta=\alpha+2$ and using the limit comparison test (comparing to $\frac{1}{n^2}$) it follows that $\sum n^{\alpha}e^{-\sqrt{n}}$ converges for all $\alpha>0$.

  • $\begingroup$ I was going to guess this was impossible before seeing this answer - great idea! I would add that unless I am missing something, any positive constant would do in place of e, and I see no reason why we couldn't generalize this to other roots of x. That would make this a very interesting class of functions. $\endgroup$
    – Will Craig
    Apr 8, 2017 at 23:29
  • $\begingroup$ Yes, $e^{-\sqrt{x}}$ was just a convenient choice, $a^{-\sqrt{x}}$ with $a>1$ would work as well. $\endgroup$ Apr 8, 2017 at 23:31

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