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It's well known that $$ \sum_{n=1}^{\infty} \frac{1}{n^{1+\varepsilon}} < \infty, \ \forall \varepsilon >0. $$

What happens if we replace $\varepsilon$ with $\varepsilon_n \downarrow 0$?

WolframAlpha says $$ \sum_{n=1}^{\infty} \frac{1}{n^{1+\varepsilon_n}} $$

converges for $$ \varepsilon_n = \frac{1}{\sqrt{\log(n+1)}} $$

and diverges for $$ \varepsilon_n = \frac{1}{\log(n+1)}. $$

So the question is: can we find a "borderline" decrease rate such that the series converge for $\varepsilon_n$ approaching zero slower than it and diverges if $\varepsilon_n$ decreases faster?

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  • $\begingroup$ yes, thanks @vadim123. $\endgroup$ – Ferdinand.kraft May 18 '13 at 2:53
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    $\begingroup$ Isn't this really just asking about the borderline of convergence for the general series $\sum a_n$, since we can just take $\epsilon_n = \log a_n / \log n - 1$? Are you aware of the divergence of, say, $\sum 1/(n \log n \log \log n \log \log \log n)$? $\endgroup$ – Erick Wong May 18 '13 at 3:04
  • $\begingroup$ @ErickWong, I see what you mean. But the general case does not have a "borderline" in the sense of vadim123's answer below. I was thinking about families of series $\sum a_n(s)$, for $s \in A \subset \mathbb{R}$. I do recognize this is not clear in the question. :-) $\endgroup$ – Ferdinand.kraft May 18 '13 at 3:17
  • $\begingroup$ That's why I asked about $\sum 1/n \log n \log \log n$ and family. It seems to me that the general case has a series of "borderlines" at different scales, starting with $1/n^s$, $1/n (\log n)^s$, $1/n \log n (\log \log n)^s$, etc. I don't see a distinction between these and the border between $\epsilon_n = 1/\log n$ and $\epsilon_n = 1/(\log n)^s$, which could be refined in a very analogous manner. $\endgroup$ – Erick Wong May 18 '13 at 4:19
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$$\int \frac{1}{x^{1+1/\ln(x)}}dx=\int\frac{dx}{x}(1/x)^{(1/\ln x)}$$

Set $u=\ln x$, then $x=e^u$ and $1/x=e^{-u}$ so the integral becomes $$\int du (e^{-u})^{1/u}=\int e^{-1}du$$

which is divergent. Replacing $1/\ln x$ with $(1/\ln x)^s$ will be divergent (resp. convergent) if $s\ge 1$ (resp. $s<1$).

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    $\begingroup$ Easier: $x^{1+1/\ln x} = ex$. $\endgroup$ – Greg Martin May 18 '13 at 5:11
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Use limit comparison test with harmonic series. We have $$\dfrac{1/n}{1/n^{1+{1/\log(1+n)}}} = n^{1/\log(1+n)} \to e$$

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