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My friend asked me this problem:

Source : Problem 35 from a 2004 book by Borwein, Bailey, and Girgensohn [1]

Determine whether the series $$\sum_{n=1}^{\infty}\frac{(\frac{2}{3}+\frac{1}{3}\cdot \sin(n))^n}{n}$$ converges.

Firstly I want to use the root test. But later I found that it just didn't work because $2/3+1/3\cdot \sin(n)$ can't be smaller than any given constant that is smaller than 1. Then I have no other ways to deal with it.

[1] Jonathan M. Borwein, David H. Bailey, and Roland Girgensohn. Experimentation in Mathematics: Computational Paths to Discovery. CRC Press, 2004.

Clement C.
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JFR
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    This is a tough one. It is proven in an article: https://arxiv.org/abs/2007.11017 – am301 Nov 27 '21 at 09:44
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    Definitely not an obvious question. Who is your friend, and are they really a friend? – Clement C. Nov 27 '21 at 10:07
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    Wait, why did this get closed? The OP has a valid and non-trivial question, and explained what they tried (the root test) and why it failed. – Clement C. Nov 27 '21 at 23:10
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    I voted to reopen the question. – user577215664 Nov 27 '21 at 23:18
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    From How to ask a good question: Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. – jjagmath Nov 28 '21 at 00:12
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    @ClementC. As an initial close voter, I think I was looking for something like a source more than an attempt here, because it was already pointed out that it was difficult. Following the editing to include the original source (as the problem textbook), I would have also voted to reopen this question. Thank you for rescuing this interesting question, along with the other reopeners. – Sarvesh Ravichandran Iyer Nov 28 '21 at 09:20
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    @teresa-lisbon : you are welcome , this post was only improved after my (now deleted) comments that it was missing references and that someone with mere knowledge of root test is highly unlikely to have friends that have access and understanding to advanced technical papers. OP showed no attempt to improve it either. Still this should be closed as it is similar to posts for math competition problems where the source is not given. Very doubtful this is the first series that root test does not work for even at the most basic courses other far simpler examples are given – jimjim Nov 29 '21 at 21:48

1 Answers1

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I summarize below the main ideas of the paper [1] whose link is in my comment above:

  1. Group the terms in the sum into "tame" and "wild" terms. The tame are defined as intergers that obey: $$ \bigg|n-\frac{\pi}2-2\pi a\bigg|\ge\frac1{n^{1/4}} $$ ($a$ integer) meaning they are "far enough" from making the sine equal to $1$. Wilds are the non-tame integers.
  2. Using the following theorem about how close $\pi$ is to rational numbers: For every integers $p,q$ such that $|q|>1$: $$ \bigg|\pi-\frac pq\bigg|>\frac1{|q|^{20}}, $$ they show that the wild numbers $W_k$ obey
    $$ W_k\ge\frac12 k^{77/76} $$ meaning they are pretty scarce.
  3. By using simple small angle expansion of the sine function they show that the sum over the tame numbers is less than the sum of $e^{-\sqrt n}$ and therefore converges.
  4. Because of their scarcity the sum over the wild numbers $W_k$ is less than or equal to twice the sum over $\frac1{k^{77/76}}$ and therefore also converges. Thus the whole sum converges.

[1] Convergence of a sinusoidal infinite series from Borwein, Bailey, and Girgensohn, Ravi B. Boppana (2020). arXiv:2007.11017

Pedro
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am301
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