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Does the following sum converge? Does it converge absolutely?

$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$$

I promise this is the last one for today:

Using Simpson's rules:

$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right) = \sum_{n=1}^\infty 2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}$$

Now, $$\left|2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}\right| \leq \frac{2}{8n² + 4n}$$

hence by the comparison test, the series converges absolutely, and hence it also converges. Is this correct?

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  • $\begingroup$ Absolutely yes. $\endgroup$ – Jack D'Aurizio Dec 26 '17 at 0:08
  • $\begingroup$ Looks just fine to me...yet the cosine's argument's denominator should be $\;\frac{4n+1}{8n^2+4n}\;$ ...like the sine's, of course. +1 $\endgroup$ – DonAntonio Dec 26 '17 at 0:09
  • $\begingroup$ Can it be studied by? $$\sin \frac{1}{2n} - \sin \frac{1}{2n+1}=\frac{1}{2n}-\frac{1}{2n+1}+o\left(\frac{1}{n^2}\right)=\frac{1}{2n(2n+1)}+o\left(\frac{1}{n^2}\right)$$ $\endgroup$ – gimusi Dec 26 '17 at 0:14
  • $\begingroup$ You have to get explicit bounds on the coefficient of the little-oh term in order to prove convergence. I think it is enough if you use $x > \sin(x) > x-x^3/6$. $\endgroup$ – marty cohen Dec 26 '17 at 7:11
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    $\begingroup$ Sure thing! Thx for the reminder $\endgroup$ – user370967 Feb 1 '18 at 22:49
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As an alternative, since:

$$\sin \frac{1}{2n} - \sin \frac{1}{2n+1}=\frac{1}{2n}-\frac{1}{2n+1}+o\left(\frac{1}{n^2}\right)=\frac{1}{2n(2n+1)}+o\left(\frac{1}{n^2}\right)$$

the series converges by limit comparison with:

$$\sum_{n=1}^\infty \frac{1}{2n(2n+1)}$$

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It's simpler still with equivalents: $$\sin \frac{1}{2n} - \sin \frac{1}{2n+1}=2\sin\frac1{4n(2n+1)}\underbrace{\cos\frac{4n+1}{4n(2n+1)}}_{\substack{\downarrow\\\textstyle1}}l\sim_\infty2\,\frac1{4n(2n+1)}\cdot 1\sim_\infty\frac1{4n^2},$$ which converges.

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Another proof: The partial sums are the same as the even partial sums of the alternating series$$\sum (-1)^n\sin \frac{1}{n}$$ which converges by the Leibniz test.

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  • $\begingroup$ Nice observation. $\endgroup$ – marty cohen Dec 26 '17 at 7:10
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It may be easier to prove a stronger result: Suppose $f$ is increasing on $[0,b].$ Let $b\ge b_1 > a_1 > b_2>a_2 > \cdots \to 0^+.$ Then

$$\sum_{n=1}^{\infty}(f(b_n)- f(a_n)) <\infty.$$

Proof: The $n$th partial sum is

$$S_n = f(b_1)-f(a_1) + f(b_2)-f(a_2)+ \cdots + f(b_n)-f(a_n)$$ $$\le\, f(b_1)-f(b_2) + f(b_2)-f(b_3)+ \cdots + f(b_n)-f(b_{n+1})$$ $$ = f(b_1)-f(b_{n+1}) \le f(b_1) - f(0).$$

Thus the sequence $S_n,$ which is increasing, is bounded above. Therefore $S_n$ converges. By definition, this implies the series converges (absolutely of course).

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Since $|\sin t|\le |t|$ we have $$|\sin(a)-\sin(b)|= 2\left|\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)\right|\le |a-b|= \frac{1}{2n}-\frac{1}{2n+1}\sim \frac{1}{n^2}$$ with $a= \frac{1}{2n}$ and $b=\frac{1}{2n+1}$

Hence, the series is absolutely convergent.

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$$|\sin(\frac{1}{2n})-\sin(\frac{1}{2n+1})| = |cos(\xi)(\frac{1}{2n}-\frac{1}{2n+1})|$$ The above is followed by mean value theorem, then $$|\sin(\frac{1}{2n})-\sin(\frac{1}{2n+1})| \leq \frac{1}{2n}-\frac{1}{2n+1}$$ So the series is absolutely convergent.

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