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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.

Blue
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James Warthington
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1 Answers1

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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 $x$

lab bhattacharjee
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  • If the sum of series is divergent, what can we say about the convergence of its constituents? That one of them must be divergent? – James Warthington Feb 10 '21 at 04:16
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    @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 $x$     – lab bhattacharjee Feb 10 '21 at 04:36