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Summing a trigonometric series $$\displaystyle\sum_{n=0}^{+\infty}\dfrac{\cos(nx)}{\cos^{n}(x)}$$

This is just a curious tribute that I wish to dedicate to Student-A Level.

Let $C=1+\dfrac{\cos(x)}{\cos(x)}+\dfrac{\cos(2x)}{\cos^{2}(x)}+\dfrac{\cos(3x)}{\cos^{3}(x)}+...$

Let $S=\dfrac{\sin(x)}{\cos(x)}+\dfrac{\sin(2x)}{\cos^2(x)}+\dfrac{\sin(3x)}{\cos^3(x)}+...$

$C+iS=1+ \dfrac{\cos(x)+i\sin(x)}{\cos(x)}+\dfrac{\cos(2x)+i\sin(2x)}{\cos^2(x)}+\dfrac{\cos(2x)+i\sin(3x)}{\cos^3(x)}+...$

We have the geometric series:

$1+\left(\dfrac{e^{ix}}{\cos(x)}\right)+\left(\dfrac{e^{ix}}{\cos(x)}\right)^2+\left(\dfrac{e^{ix}}{\cos(x)}\right)^3+\left(\dfrac{e^{ix}}{\cos(x)}\right)^4...$

Using the geometric summing formula for infinite series

$\dfrac{\cos(x)}{\cos(x)+e^{ix}}=\dfrac{\cos(x)}{2\cos(x)+i\sin(x)}$

The real part is what I don't know.

Is this a valid derivation, or it is just rubbish?

Using this method, I cannot sum the infinte series, which is $C$, am I right?

I also have no clue whether it converges for diverge.

I hope to get a divergent series after doing this stuff to have one more example.

Wolfie won't sum this series.

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  • $\begingroup$ @JamesWarhington If |cos x| < 1, then $cos^n x \to 0$ as $n \to \infty$ whereas the same is not true for the numerator. It can also be shown using the fact that the r of your complex geometric series is |sec x| which is greater than 1. $\endgroup$ Feb 10, 2021 at 4:21
  • $\begingroup$ Both series $\cos(nx)$ and $\cos^{n}(x)$ are divergent. $\endgroup$ Feb 10, 2021 at 4:22
  • $\begingroup$ I know $cos(nx)$ is divergent because I asked it before. $\endgroup$ Feb 10, 2021 at 4:24
  • $\begingroup$ @JamesWarhington IF |cos x| < 1 then the $cos^n x$ converges to 0 for sure. $\endgroup$ Feb 10, 2021 at 4:26

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We can of course rationalize the denominator and extract the real part

$$\dfrac{\cos x}{2\cos x+i\sin x}=\dfrac{\cos x(2\cos x-i\sin x)}{(2\cos x)^2+(\sin x)^2}$$

But unfortunately the infinite sum does not converge as the common ratio $$=\left|\dfrac{e^{ix}}{\cos x}\right|=|\sec x|\ge1$$ for real $x$

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  • $\begingroup$ If the sum of series is divergent, what can we say about the convergence of its constituents? That one of them must be divergent? $\endgroup$ Feb 10, 2021 at 4:16
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    $\begingroup$ @JamesWarthington, Correct. More explicitly, $$2\sum_{n=0}^\infty\dfrac{\cos nx}{\cos^nx}=\sum_{n=0}^\infty\left(\dfrac{e^{ix}}{\cos x}\right)^n+\sum_{n=0}^\infty\left(\dfrac{e^{-ix}}{\cos x}\right)^n$$ Here both series are divergent for real $x$ $\endgroup$ Feb 10, 2021 at 4:36

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