Infinite series of trigonometric ratios

Question

The question is to compute: $$(1+\cos A)+2(1+\cos A)^2 + 3(1+\cos A)^3+\ldots = \sum_{k=1}^{\infty}k(1+\cos A)^k.$$

I tried by setting $1+\cos A=y$, then the serie becomes $$y+2y^2+3y^3+\ldots = \sum_{k=1}^{\infty}ky^k$$
It's not a geometric progression as the coefficients are not in the series.

How can I go further?
Hints are welcome.

$y+2y^2+3y^3+...=y(1+2y+3y^2+...)$ Now, what is inside parentheses ? — Claude Leibovici, Aug 23 '15 at 16:12
Differentiate the geometric progression $\sum_n y^n$ with respect to $y$. — Start wearing purple, Aug 23 '15 at 16:13

score 5 · Accepted Answer · answered Aug 23 '15 at 16:16

Using Ratio test,

$$\lim_{n\to\infty}\dfrac{(n+1)(1+\cos A)^{n+1}}{n(1+\cos A)^n}=\left(1+\cos A\right)\lim_{n\to\infty}\left(1+\dfrac1n\right)=1+\cos A$$

So, the series can only converge if $|1+\cos A|<1\iff 1+\cos A<1\iff\cos A<0$

For $|y|<1,$ let $$S_n=\sum_{r=1}^nr y ^r$$

$$yS_n=\sum_{r=1}^nry^{r+1}$$

$$(1-y)S=y+y^2+y^3+\cdots=\dfrac y{1-y}$$

score 0 · Answer 2 · edited Apr 13 '17 at 12:21

0

Theorem: $$1+2x+3x^2+4x^3+\dotsb=\frac1{(1-x)^2}$$ Proof: Proof without words

Now, multiply everything by $x$ and let $x=1+\cos A$.

For more info see this thread.

edited Apr 13 '17 at 12:21

Community

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answered Aug 23 '15 at 16:38

Akiva Weinberger

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Infinite series of trigonometric ratios

2 Answers2