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$\sum_\limits{x=0}^∞ cos(x)$

= Re( $\sum_\limits{x=0}^∞ exp(ix)$)

=Re ($\frac{1}{1 - \exp(ix)}$)

= $\frac{1}{1 - \cos(x)}$

I know this is wrong but how would I use this technique to calculate that sum?

Jose M Serra
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jdhokia
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    Two wrongs… 1. The series $\sum_x\exp(ix)$ is divergent. 2. $\operatorname{Re}(1/z)\neq1/\operatorname{Re}(z)$… don't make a right. – gniourf_gniourf Dec 22 '16 at 15:01
  • Why is $\Re\left(\frac{1}{1-exp(ix)}\right)=\frac{1}{1-\cos(x)}$? In general, $\Re\left(\frac{1}{a+bi}\right)\not=\frac{1}{a}$. More generally, $\Re(z_1z_2)\not=\Re(z_1)\Re(z_2)$. – Michael Burr Dec 22 '16 at 15:01
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    Let $z\neq 0$ a complex number, then you can calculate $\Re\frac{1}{z}$, from $$\Re\frac{1}{z}=\Re\frac{\bar{z}}{z\bar{z}}=\Re\frac{\bar{z}}{|z|^2}.$$ Thus that is $\frac{\Re \bar{z}}{|z|^2}$. –  Dec 22 '16 at 15:24

3 Answers3

3

Note that

$$\sum_{n=0}^\infty r^n=\frac1{1-r}\text{ iff }|r|<1$$

In this case, we have $|e^i|=1$, but $1$ is not less than $1$.

This is like saying that

$$\sum_{n=0}^\infty(-1)^n=\frac12$$

Indeed, your sum diverges by the term test. For a sum to approach some value, you have to be adding on increasingly smaller values, but $\cos(x)$ does not approach $0$.

In the context of regularizing the sum, you could take a power series regularization:

$$\lim_{k\ \uparrow\ 1}\sum_{n=0}^\infty k^n\cos(n)=\lim_{k\ \uparrow\ 1}\Re\left(\frac1{1-ke^i}\right)=\frac12$$

Simply Beautiful Art
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2

Instead of attempting to sum the divergence series, we will evaluate a modification of that series. Let $-1<\lambda<1$. Then, we have

$$\begin{align} \sum_{n=0}^\infty \lambda^n \cos(nx)&=\text{Re}\left(\sum_{n=0}^\infty (\lambda e^{ix})^n\right)\\\\ &=\text{Re}\left(\frac{1}{1-\lambda e^{ix}}\right)\\\\ &=\text{Re}\left(\frac{1-\lambda e^{-ix}}{1+\lambda^2-2\lambda \cos(x)}\right)\\\\ &=\frac{1-\lambda \cos(x)}{1+\lambda^2-2\lambda \cos(x)}\tag 1 \end{align}$$

While the expression in $(1)$ is valid for $\lambda\in (-1,1)$, the right-hand side can be evaluated at $\lambda =\pm1$ with the result $\frac12$ for all $x\in (0,\pi)$.

As a side note, we have for $\lambda \in (-1,1)$

$$\sum_{n=0}^\infty \lambda^n \sin(nx)=\frac{\lambda \sin(x)}{1+\lambda^2-2\lambda \cos(x)}$$

for which the right-hand side at $\lambda =1$ is $\frac{\sin(x)}{2(1-\cos(x))}=\frac12 \cot(x/2)$ for $x\in (0,\pi)$

Mark Viola
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2

we have :$\lim_{n\to+\infty}\cos(n)\neq 0.$

if not, we will have

$$\lim_{n\to+\infty}\cos(2n)=0$$

$$\lim_{n\to+\infty}(2\cos^2(n)-1)=-1$$

thus $\sum \cos(n)$ diverges and we cannot write $\sum_{x=0}^{+\infty}\cos(x)$.

hamam_Abdallah
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