How Do p^n Cos(nx) and p^n Sin(nx) Converge?

[ognik]

They ask for both $ \sum_{n=0}^{\infty} p^n Cos nx, also \: p^n Sin (nx) $

I'm thinking De Moivre so $$\sum_{n=0}^{\infty}p^n (e^{ix})^n = \sum_{n=0}^{\infty} p^n(Cos x + i Sin x)^n= \sum_{n=0}^{\infty} (pCos x + ip Sin x)^n$$

I also tried a geometric series with a=1, $r=pe^{ix}$

But those won't work out with the limit of $\infty$, so any hints please?

 

[Prove It]

They ask for both $ \sum_{n=0}^{\infty} p^n Cos nx, also \: p^n Sin (nx) $

I'm thinking De Moivre so $$\sum_{n=0}^{\infty}p^n (e^{ix})^n = \sum_{n=0}^{\infty} p^n(Cos x + i Sin x)^n= \sum_{n=0}^{\infty} (pCos x + ip Sin x)^n$$

I also tried a geometric series with a=1, $r=pe^{ix}$

But those won't work out with the limit of $\infty$, so any hints please?

What makes you think the limit is $\displaystyle \begin{align*} \infty \end{align*}$? The geometric series is convergent where $\displaystyle \begin{align*} |r| < 1 \end{align*}$, so where

$\displaystyle \begin{align*} \left| p\,\mathrm{e}^{\mathrm{i}\,x} \right| &< 1 \\ \left| p \right| \left| \mathrm{e}^{\mathrm{i}\,x} \right| &< 1 \\ \left| p \right| \cdot 1 &< 1 \\ \left| p \right| &< 1 \end{align*}$

As long as you have $\displaystyle \begin{align*} \left| p \right| < 1 \end{align*}$ the series is convergent.Also $\displaystyle \begin{align*} \mathrm{e}^{\mathrm{i}\,\theta} \equiv \cos{ \left( \theta \right) } + \mathrm{i}\sin{ \left( \theta \right) } \end{align*}$ is Euler, not DeMoivre...

 

[ognik]

Thanks Proove it, I was thinking $s_n = \frac{1-r^n}{1-r}$, but if |p| < 1 then |r| < 1 and $s_n = \frac{1}{1-r} = \frac{1}{1-pe^{ix}}$ and with Euler's help it all works out fairly easily :-)