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How to show that, for any fixed constant $a\in(0,1)$, the series $$ \sum_{n=1}^{\infty}(-1)^n \sin\left(\frac{a}{n}\right). $$ is convergent yet not absolutely convergent.

My idea is to express sin(x) as series but I don't know if that makes sense since $\sin(a/n)$ is in series?

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  • $\begingroup$ $1<0$? $\quad \quad$ $\endgroup$
    – Hayden
    Oct 30, 2014 at 16:48
  • $\begingroup$ It's viceversa! @Hayden $\endgroup$
    – MM13 J044
    Oct 30, 2014 at 16:50

4 Answers 4

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Hint: Fix $a\in(0,1)$, Then $$(-1)^n\sin \frac{a}{n} = (-1)^n\left(\frac{a}{n} - \frac{a^3}{6n^3} + o\!\left(\frac{1}{n^3}\right)\right) = a_n + b_n$$ where $a_n=\frac{(-1)^n a}{n}$ and $b_n = \frac{(-1)^n a^3}{6n^3} + o\!\left(\frac{1}{n^3}\right)$. Now, $\sum a_n$ is conditionally convergent, $\sum b_n$ absolutely convergent; what does that imply for the original series?

As a remark, this is a general technique worth knowing: do a Taylor expansion until you get to a term whose series (in particular including the remainder) is absolutely convergent — so that it's easy to deal with this one. Then argue the series of the first other terms are conditionally convergent.

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$\sin(\frac{a}{n})>0$ and decreases to 0, hence by alternating test, the series converges.

Since $|\sin(x)|>\frac{2x}{\pi}$ for $|x|<\frac{\pi}{2}$, so $|\sin(\frac{a}{n})|>\frac{2a}{\pi n}$, hence by comparison test, it's not absolutely convergent.

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  • $\begingroup$ More precisely, $\sin\frac an$ is eventually strictly decreasing $\endgroup$ Oct 30, 2014 at 16:58
  • $\begingroup$ @HagenvonEitzen Since $a\in (0,1)$, isn't it always strictly decreasing? $\endgroup$
    – John
    Oct 30, 2014 at 17:01
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Hint: $\sum |(-1)^n\sin (a/n)|=\sum |\sin(a/n)|\sim \sum |a/n|$... For conditional convergence, use Leibniz criterion.

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We shall use the following facts:

Fact A. $f(x)=\sin x$ is positive and strictly increasing in $(0,\pi/2)$.

Fact B. If $x\in (0,\pi/2)$, then $$ \frac{2 x}{\pi} < \sin x<x. $$

Fact C. If $a_1\ge a_2\ge \cdots \ge a_n$ and $a_n\to 0$, then the series $\sum_{n=1}^\infty (-1)^na_n$ converges conditionally. (Possibly not absolutely.)

The series converges conditionally. Clearly the sequence $\{\sin \left(\frac{a}{n}\right)\}_{n\in\mathbb N}$ satisfies the requirements of Fact C, as $$ \sin \left(\frac{a}{n}\right)>\sin \left(\frac{a}{n+1}\right), $$ due to Fact A. Hence $\sum_{n=1}^\infty \sin \left(\frac{a}{n}\right)$ converges. (At least conditionally.)

The series does not converge absolutely. Due to Fact B $$ \sin \left(\frac{a}{n}\right) \ge \frac{2\pi a}{n}, $$ and hence $$ \sum_{n=1}^\infty \left\lvert\,\sin \left(\frac{a}{n}\right)\right\rvert\ge \sum_{n=1}^\infty\frac{2a}{\pi n}=\infty. $$

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