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Determine convergence of $$\sum_{n=1}^{\infty} \left(\cos{\frac{2}{n}}-\cos{\frac{4}{n}}\right)$$

In the answer, it says

$$\cos{\frac{2}{n}}-\cos{\frac{4}{n}} = 2\sin{\frac{3}{n}}\sin{\frac{1}{n}} \le 2\cdot \frac{3}{n} \cdot \frac{1}{n} = \frac{6}{n^2}$$

But how do I get the above trig substitution? I guess removing the fractions, I will get $\cos{x}-\cos{2x}=2\sin{(2x-1)}\sin{(x-1)}$ ... probably this is wrong, but how do I get that?

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See Product-to-sum and sum-to-product identities at Wikipedia. –  Martin Sleziak May 1 '12 at 6:40

2 Answers 2

up vote 4 down vote accepted

The result follows from the Addition Law for Cosines.

We have $\cos(x+y)=\cos x\cos y-\sin x\sin y$ and $\cos(x-y)=\cos x\cos y+\sin x\sin y$. Subtract. We get $$\cos(x-y)-\cos(x+y)=2\sin x\sin y.$$ Let $x-y=\frac{2}{n}$ and $x+y=\frac{4}{n}$. Solve for $x$ and $y$. We get $x=\frac{3}{n}$ and $y=\frac{1}{n}$.

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Oh, now how do I get $2\sin{\frac{3}{n}}\sin{\frac{1}{n}} \le 3\cdot \frac{3}{n}\cdot \frac{1}{n}$? Is it squeeze theorm? But won't I have $$-1 \le \sin{\frac{3}{n}} \le 1$$ $$-2 \le 2\sin{\frac{3}{n}} \le 2$$ $$-2 |\sin{\frac{1}{n}}| \le 2\sin{\frac{3}{n}}\sin{\frac{1}{n}} \le 2|\sin{\frac{1}{n}}|$$ then as $n\rightarrow 0$, by limit goes to 0? So convergent? –  Jiew Meng May 1 '12 at 7:45
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@JiewMeng: No. You need to prove the convergence of a series, and it is not sufficient (but it is necessary) for the sequence of individual terms to converge to zero to conclude that the series converges. You need to use a suitable dominating series here. Also, you need to remember a result telling which is bigger $\sin x$ or $x$, when $x$ is positive? A reference to a theorem in your book is absolutely needed for full credit. –  Jyrki Lahtonen May 1 '12 at 7:55
    
@JyrkiLahtonen, thanks, but how do i get $2\sin{\frac{3}{n}}\sin{\frac{1}{n}} \le \color{blue}2 \cdot \frac{3}{n}\cdot \frac{1}{n}$ then?. highlighted typo in last comment, should be 2 instead of 3 –  Jiew Meng May 1 '12 at 8:03
    
@JiewMeng: I think I already answered that? You must have done that bit earlier in your class for the hint given in your book to make sense! If $a>b>0$ and $c>d>0$, then $a\cdot c > b\cdot d$? –  Jyrki Lahtonen May 1 '12 at 8:08
    
hmm, actually that was the answer, hmm... I'm terrible at maths ... need to think a bit –  Jiew Meng May 1 '12 at 8:27

I suggest another way:

$\cos\left(\frac{2}{n}\right)-\cos\left(\frac{4}{n}\right)=-\left(1-\cos\left(\frac{2}{n}\right)\right)+1-\cos\left(\frac{4}{n}\right)\sim -\frac{4}{2n^2}+\frac{16}{2n^2}=\frac{6}{n^2}$ and $\sum \frac{6}{n^2}$ converges.

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