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If series $\sum_{n=1}^{\infty} a_{n}$ converges, prove series $\sum_{n=1}^{\infty} \sin(a_{n})$ converges too.

Is this a series function problem or something related to?

I tried this:

If $a_{n} \ge 0 $ for all $n$, then $| \sin (a_{n})| \le a_{n}$, on the other hand $\sin(x)$ is continuous function in $[0,x]$ and differentiable in $(0,x)$ then exist $c \in (0,x)$ and $$(x-0)\cos (c)= \sin(x)-\sin(0)$$ then $$x\cos(c)= \sin(x)$$

but, $|\sin(x)|=|x\cos(c)|= |x| |\cos(c)| \le |x|$, thus $|\sin (x)| \le |x|$.

We know, $|\sin (a_{n})| \le |a_{n}| $ and $\sum_{n=1}^{\infty} a_{n}$ converges, then $\sum_{n=1}^{\infty} |\sin(a_{n})|$ converges. Therefore $\sum_{n=1}^{\infty} \sin(a_{n})$ converges.

But this proof use ${a_{n}}$ positive and in the original problem I don't have this hypotesis.

Marinela
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  • I think you have a typo? Is it $\sin$ ? – jl00 May 26 '20 at 20:12
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    Might just be a language issue. Some books written in other languages might write sen for sin. Anyways, Mari, can you show us any kind of working you may have done, as best as you can? – Abhijeet Vats May 26 '20 at 20:14
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    Is, by any chance, given that $;{a_n};$ is a positve sequence (i.e., is the series a positive one?) – DonAntonio May 26 '20 at 20:18
  • @AbhijeetVats yes, it was my mistake, my native language is spanish and I forgot the difference, I edit the post and add my work :) – Marinela May 26 '20 at 20:40
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    ,@DonAntonio no, it's not, the problem just say that. – Marinela May 26 '20 at 20:40
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    Without the non-negativity assumption, the conclusion need not hold. – Daniel Fischer May 26 '20 at 22:39
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    @DanielFischer: That surprised me! But the proof is not difficult. – TonyK May 26 '20 at 23:27
  • It is trivial to see that if one takes $a_n=\frac{1}{n^{1/3}}, n=3k+1, a_n= -\frac{1}{2n^{1/3}}, n=3k+2, n=3k+3$, $\sum a_n$ converges, but $\sum \sin (a_n) \to -\infty$ as the $-x^3$ term in the taylor series dominates (the higher terms are absolutely convergent) and the linear term sums to a finite sum by construction – Conrad May 26 '20 at 23:29
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    @DanielFischer It might be good to make your comment an answer, so that (1) it can be accepted to prevent the software form treating the question as unsolved and (2) people don't keep trying to prove the contrary. – Andreas Blass May 27 '20 at 01:26

3 Answers3

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This is a "Community Wiki" answer recording a comment by Daniel Fischer under the question. The comment provides a link The set of functions which map convergent series to convergent series to a proof that the result in question is false in general, though certainly true when the $a_n$ are non-negative. My reason for writing this answer is that comments can vanish more easily than answers and can probably also be more easily overlooked.

Andreas Blass
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    Neat. If I understand that proof properly, you can construct an explicit counterexample by considering the telescoping sum given by taking the triples $2/n, -1/n, -1/n$, repeating each such triple several times, and putting them all into a series. The $n$th triple should be repeated about $n^3$ times, corresponding to the nonlinearity $\sin(2x) - \sin(x) - \sin(x) \approx -x^3$. – Jair Taylor May 28 '20 at 01:05
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If $(a_n)_n$ is assumed non-negative (or non-positive), this follows from the comparison theorem as $0 \leq |\sin a_n| \leq a_n$.

However, the conclusion is false otherwise. Here is a counterexample: we are given in this answer an explicit sequence $(a_n)_n$ such that

  • $\sum_n a_n$ converges
  • $\sum_n a_n^3$ diverges
  • $\sum_n a_n^4$ converges (by inspection of the explicit sequence given)

Specifially: for all $n\geq 1$, $$ a_{3n-2} = \frac{1}{n^{1/3}}, \quad a_{3n-1} = a_{3n} = -\frac{1}{2n^{1/3}} $$

In particular, clearly, $\lim_{n\to\infty}a_n =0$. Since $\sin x = x-\frac{x^3}{6} + O(x^4)$, we get $$ \sum_n \sin(a_n) = \sum_n a_n -\frac{1}{6}\sum_n a_n^3 + O\left(\sum_n a_n^4\right) $$ (the use of $O(\cdot)$ here is OK, as we deal with an absolutely convergent series at that point). But $\sum_n a_n, \sum_n a_n^4$ are convergent (convergent and absolutely convergent, respectively), while $\sum_n a_n^3$ isn't: so the RHS diverges. So the LHS must diverge too.

Clement C.
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  • Note: The only non-obvious part is the "(by inspection of the explicit sequence given)". But it is because in that explicit example, $\sum_{n=1}^\infty a_n^4 = \sum_{k=1}^\infty (\frac{1}{k^{4/3}}+ 2\cdot \frac{1}{2^4k^{4/3}}) < \infty$. – Clement C. May 28 '20 at 00:31
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Write $ \sum_n \sin(a_n) = \sum_n a_n + \sum_n a_n(\frac{\sin(a_n)-a_n}{a_n}) $. Since $ \sum_n a_n $ converges, $ \lim_n a_n = 0 $. Thus, for $ n $ large enough, by a Taylor series expansion, $ |\frac{\sin(a_n)-a_n}{a_n}| = O(a_n^2) $. Hence, by the triangle inequality, $ |\sum_n \sin(a_n)| \leq | \sum_n a_n | + O(\sum_n |a_n|^3) $. Finally, we use the fact that if $ a_n $ is nonnegative, then $ \sum_n |a_n|^3 $ converges whenever $ \sum_n a_n $ converges.

mr_snazzly
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    You make what seems to be a series of claims leading to a proof. Then you say that of course these claims are only valid if the $a_n$ are non-negative. So you lied! Can you at least tell us where your chain of inferences beaks down? – TonyK May 27 '20 at 20:21
  • $ \sum_n a_n $ converges $ \implies $ $ \sum_n |a_n|^3 $ converges requires the $a_n$ to be nonnegative. – mr_snazzly May 27 '20 at 20:57
  • But if $a_n$ is non-negative, then $|\sin a_n|\leq a_n$ is enough to conclude... – Clement C. May 28 '20 at 00:11
  • True, though my solution shows that the conclusion is true for general $ a_n $ as long as $ \sum_n a_n $ and $ \sum_n |a_n|^3 $ both converge. – mr_snazzly May 28 '20 at 01:24