Emanuel Keith

2022-06-21

Finding the value of trignometric series ${\mathrm{\Sigma}}_{0}^{\mathrm{\infty}}{\displaystyle \frac{\mathrm{cos}nx}{{3}^{n}}}$

Bruno Hughes

Beginner2022-06-22Added 25 answers

Hint. One may write, for $x\in \mathbb{R}$,

$\frac{\mathrm{cos}nx}{{3}^{n}}}=\text{Re}{\left({\displaystyle \frac{{e}^{ix}}{3}}\right)}^{n}=\text{Re}\phantom{\rule{mediummathspace}{0ex}}{z}^{n$

with $z={\displaystyle \frac{{e}^{ix}}{3}}$ then one may recall that

$\sum _{n=0}^{\mathrm{\infty}}{z}^{n}=\frac{1}{1-z},\phantom{\rule{1em}{0ex}}|z|<1.$

$\frac{\mathrm{cos}nx}{{3}^{n}}}=\text{Re}{\left({\displaystyle \frac{{e}^{ix}}{3}}\right)}^{n}=\text{Re}\phantom{\rule{mediummathspace}{0ex}}{z}^{n$

with $z={\displaystyle \frac{{e}^{ix}}{3}}$ then one may recall that

$\sum _{n=0}^{\mathrm{\infty}}{z}^{n}=\frac{1}{1-z},\phantom{\rule{1em}{0ex}}|z|<1.$