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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](http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities) at Wikipedia.2012-05-01

2 Answers 2

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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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    For positive $x$, we have $\sin x \le x$. Proof: Let $f(x)=x-\sin x$. We have $f(0)=0$. Note that $f'(x)=1-\cos x \ge 0$, so $f(x)$ is non-decreasing, and is therefore $\ge 0$ for all $x \ge 0$. It follows that $\sin x \le x for all $x \ge 0$. So \sin(1/n) \le $1/n$ (actually, less), and $\sin(3/n) \le 3/n$.2012-05-02
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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.