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This is the generalized form for a question I got on the midterm review where I had to state why for $a=5$, the sequence failed to converge.

$$ \mbox{Does the series}\quad \sum_{n=1}^{\infty} \sin^{n}\left(n\pi \over a\right)\quad \mbox{converge when}\ a \in \mathbb{N}\ ? $$ and how does the value of $a$ impact the series' convergence ?.

After applying the Root Test, I got that the value lies between $\left(0,1\right)$. Since $L = 1$ means that the test is inconclusive, I do not know how to continue.

Richard D
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    Where did this question come from? – Mike Oct 16 '24 at 18:26
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    |sin(x)| <= 1 so I think by inspection that this series will not converge when a=2 since successive terms in the series $\sin(n\pi/2)$ will repeat the sequence 0,1,0,-1, forever. I think the same convergence problem arises whenever a is an even integer. – Martin Brown Oct 16 '24 at 18:30
  • @Mike This is a practice question for my upcoming midterm. – Richard D Oct 16 '24 at 18:31
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    @Steven, The sequence for $a=4$ produces a $1$ for $n=4k+2$, $0$ for $n=4k$ and $2$ bigger positive terms than $2$ negative terms for $n = 4k+1$ and $n =4k+3$. Net value is positive after every $8$ terms. – Srini Oct 16 '24 at 18:55
  • It is not the magnitude of $a$ that matters most. You might want to look at pisot numbers and the alike. – mick Oct 16 '24 at 21:59
  • This reminds me a lot of this question, the solution fo which was extremely complicated. – K.defaoite Oct 16 '24 at 22:32
  • this is a weird problem to have for midterm preparation unless $a$ is specified to be rational when the answer is complete as the series converges iff in lowest terms $a$ has odd numerator (since the terms are periodic so they form a few geometric series, the only bad case is when you get $\pm 1$ and that is easily seen to happen precisely when the numerator of $a$ is even in lowest terms); however it is easy to construct uncountable many examples of irrational $a$ for which the series is divergent by taking $1/a$ decimal sequences that are very sparse – Conrad Oct 16 '24 at 23:35
  • I made a mistake, a should be a natural number. – Richard D Oct 16 '24 at 23:46

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If $a$ is required to be a positive integer as OP stated in comments, the answer is the sum $\sum_{n=0}^{\infty} \Big(\sin \Big(\frac{n\pi}{a} \Big)\Big)^n$ does not converge.

Indeed, if this sum did converge, then there would be for all $\epsilon >0$ an integer $n'$ such that the relation $\Big|\Big(\sin\Big(\frac{2\pi n}{a}\Big)\Big)^n \Big|$ $\le \epsilon$ holds for all $n \ge n'$. But there is no such integer $n'$. Indeed, for any such integer $n'$, the equation $\Big|\Big(\sin \Big(\frac{n\pi}{a}\Big)\Big)^n\Big| = 1$ is true for at least one $n \ge n'$, namely, every $n$ that is an odd multiple of $a$.

Mike
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