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Does anyone know how to determine with proof whether the series $$\sum_{n=1}^\infty\frac{1}{n^{2+\cos(2\pi\ln(n)) }}$$ converges?

David
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  • Is there an integer value of $n$ for which the argument causes the cosine to become $-1$? – imranfat Mar 10 '16 at 00:05
  • @imranfat There are certainly integers for which the cosine is close to $-1$. The question is how many such integers there are, with the cosine how close to $-1$... – David C. Ullrich Mar 10 '16 at 00:08
  • Yes, close to $-1$, however, the exponent of $n$ in the denom is then still greater than $1$, right? – imranfat Mar 10 '16 at 00:09
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    @imranfat Note $\sum 1/n^{1+1/n} = \infty.$ – zhw. Mar 10 '16 at 00:11
  • @zhw.Yes, true (good point actually) but is there a way perhaps to get "around" this "1/n" term that the exponent grows "less" fast? The denom does not approach 1/n I think – imranfat Mar 10 '16 at 00:14
  • @imranfat Like I said, that's the question - how close to $1$ is the exponent, for how many values of $n$. – David C. Ullrich Mar 10 '16 at 00:15
  • Hmmm, I have to admit that ZHW threw a little "dirt" on my idea... – imranfat Mar 10 '16 at 00:17
  • @imranfat I haven't thought about it much, but if I'm not missing something obvious this is much trickier than the typical "does this series converge?" question we see here. – David C. Ullrich Mar 10 '16 at 00:18
  • You are not missing something obvious at all. I am currently staring at a white wall... – imranfat Mar 10 '16 at 00:19
  • I upvoted the question, it is a great one. Does this question come from a Calc book or something? Just wondering.... – imranfat Mar 10 '16 at 00:21
  • Of possible interest https://www.desmos.com/calculator/dvjw2jhslh –  Mar 10 '16 at 01:00

1 Answers1

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I am not completely sure in one argument below, but I think the series diverges. Let $\delta > 0$ be small and let $\varepsilon > 0$ be such that $\cos(\pi\pm\delta)=-1+\varepsilon$. Moreover, put $\tau := \delta/(2\pi)$. For $k\in\mathbb N$ define the set $\Delta_k^\tau := (e^{k+1/2-\tau},e^{k+1/2+\tau})$. Then we have $n\in\Delta_k^\tau$ if and only if $2+\cos(2\pi\ln(n))\in [1,1+\varepsilon)$ and the series is as least as large as $$ \sum_k\sum_{n\in\Delta_k^\tau}\frac 1 {n^{1+\varepsilon}}. $$ And now, I am not 100% sure. I claim that $$ \sum_{n\in\Delta_k^\tau}\frac 1 {n^{1+\varepsilon}}\,\ge\,\int_{\Delta_k^\tau}\frac 1 {x^{1+\varepsilon}}\,dx. $$ If not, then twice the left guy should do. Now, $$ \int_{\Delta_k^\tau}\frac 1 {x^{1+\varepsilon}}\,dx = -\frac 1 {\varepsilon e^{\varepsilon/2}}\left(\frac 1{e^{\varepsilon\tau}} - \frac 1{e^{-\varepsilon\tau}}\right)\cdot e^{-\varepsilon k} = \frac 2 {\varepsilon e^{\varepsilon/2}}\sinh(\varepsilon\tau)e^{-\varepsilon k}. $$ Summing over $k$, we get $$ \frac 2 {\varepsilon e^{\varepsilon/2}}\sinh(\varepsilon\tau)\frac 1 {1-e^{-\varepsilon}} = \frac 1 \varepsilon \cdot\frac{\sinh(\varepsilon\tau)}{\sinh(\varepsilon/2)} = \frac 1 \pi\cdot\frac{\delta}{\varepsilon}\cdot\frac{\sinh(\varepsilon\tau)}{\varepsilon\tau}\cdot\frac{\varepsilon/2}{\sinh(\varepsilon/2)}. $$ Now, we let $\delta\to 0$. Then, of course, also $\tau\to 0$ and $\varepsilon\to 0$. So, the last two factors tend to one. But $\delta/\varepsilon\to\infty$. Indeed, we have $\cos(\delta) = 1-\varepsilon$, so $\delta = \arccos(1-\varepsilon)$ and $$ \lim_{x\downarrow 0}\frac{\arccos(1-x)}{x} = \infty. $$ This (hopefully) shows that the series diverges.

Friedrich Philipp
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