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Question Number 3650 by prakash jain last updated on 18/Dec/15

$$\mathrm{If}{a}_n\mathrm{is}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{all}\mathrm{primitive}{n}^{th}\mathrm{root}\mathrm{of}$$$$\mathrm{unity}.$$$$\mathrm{Does}\underset{n=1}{\sum ^{\mathrm{\infty }}}{a}_n\mathrm{converge}?$$$$\mathrm{An}{n}^{th}\mathrm{root}\mathrm{of}\mathrm{unity},\mathrm{say}z,\mathrm{is}\mathrm{primitive}\mathrm{if}\mathrm{it}\mathrm{is}\mathrm{not}$$$$k^{th}\mathrm{root}\mathrm{of}\mathrm{unity}\mathrm{where}k<n.$$$$or{z}^n=1\mathrm{and}{z}^k\ne 1(\mathrm{k}=1,2,3,..,n-1)$$$$a_1=1$$$$a_2=-1(2^{nd}\mathrm{root}\mathrm{of}\mathrm{unity}\mathrm{are}1\mathrm{and}-1,1\mathrm{is}\mathrm{not}$$$$\mathrm{primitive})$$

Answered by Yozzii last updated on 18/Dec/15

$$Letp\left(n\right)betheprimitiverootsofthe$$$$equation{z}^n=1and{a}_n=\mathrm{\Sigma }p\left(n\right).$$$$n=1\Rightarrow z=1\Rightarrow p\left(1\right)=1\Rightarrow a_1=1.$$$$n=2\Rightarrow z^2=1\Rightarrow z=\pm 1\Rightarrow p\left(2\right)=-1\Rightarrow a_2=-1.$$$$n=3\Rightarrow z^3=1=e^{2\pi ir}(-1⩽r⩽1)$$$$\Rightarrow z=e^{\frac{2\pi ir}{3}}\Rightarrow p\left(3\right)=e^{\pm \frac{2\pi i}{3}}\Rightarrow a_3=2cos\frac{2\pi }{3}=-1$$$$n=4\Rightarrow z^4=1\Rightarrow z=\pm 1,\pm i\Rightarrow p\left(4\right)=\pm i\Rightarrow a_4=0$$$$n=5\Rightarrow z^5=e^{2\pi ir}(-2⩽r⩽2)$$$$\Rightarrow z=e^{\frac{2\pi ir}{5}}=1,e^{\pm \frac{2\pi i}{5}},e^{\pm \frac{4\pi i}{5}}\Rightarrow p\left(5\right)=e^{\pm \frac{2\pi i}{5}},e^{\pm \frac{4\pi i}{5}}$$$$\Rightarrow a_5=2(cos\frac{2\pi }{5}+cos\frac{4\pi }{5})$$$$a_5=2\left\{2cos\frac{\frac{4\pi +2\pi }{5}}{2}cos\frac{\frac{4\pi -2\pi }{5}}{2}\right\}$$$$a_5=4cos\frac{3\pi }{5}cos\frac{\pi }{5}=-1$$$$n=6\Rightarrow z^6=e^{2\pi ir}(-2⩽r⩽3)$$$$z=e^{\frac{\pi ir}{3}}\Rightarrow z=\pm 1,e^{\pm \frac{\pi i}{3}},e^{\pm \frac{2\pi i}{3}}\Rightarrow p\left(6\right)=e^{\pm \frac{\pi i}{3}}$$$$\Rightarrow a_6=2cos\frac{\pi }{3}=1$$$$n=7\Rightarrow z^7=e^{2\pi ir}(-3⩽r⩽3)$$$$\Rightarrow z=e^{\frac{2\pi ir}{7}}\Rightarrow z=1,e^{\frac{\pm 2\pi i}{7}},e^{\frac{\pm 4\pi i}{7}},e^{\frac{\pm 6\pi i}{7}}$$$$\Rightarrow a_7=2(cos\frac{2\pi }{7}+cos\frac{4\pi }{7}+cos\frac{6\pi }{7})=-1$$$$n=8\Rightarrow z^8=e^{2\pi ir}\Rightarrow z=e^{\frac{\pi ir}{4}}(-3⩽r⩽4)$$$$\Rightarrow z=\pm 1,e^{\pm \frac{3\pi i}{4}},\pm i,e^{\pm \frac{\pi i}{4}}$$$$\Rightarrow a_8=2(cos\frac{3\pi }{4}+cos\frac{\pi }{4})=0.$$$$n=9\Rightarrow z=e^{\frac{2\pi ir}{9}}(-4⩽r⩽4)$$$$\Rightarrow z=1,e^{\pm \frac{2\pi i}{9}},e^{\pm \frac{4\pi i}{9}},e^{\pm \frac{2\pi i}{3}},e^{\pm \frac{8\pi i}{9}}$$$$\Rightarrow a_9=2(cos\frac{2\pi }{9}+cos\frac{4\pi }{9}+cos\frac{8\pi }{9})=0$$$$n=10\Rightarrow z=e^{\frac{\pi ir}{5}}(-4⩽r⩽5)$$$$\Rightarrow z=\pm 1,e^{\pm \frac{\pi i}{5}},e^{\pm \frac{2\pi i}{5}},e^{\pm \frac{3\pi i}{5}},e^{\pm \frac{4\pi i}{5}}$$$$\Rightarrow a_{10}=2(cos\frac{\pi }{5}+cos\frac{3\pi }{5})=1$$$$$$$$Iconjectthatfornbeingprime,$$$$a_n\ne 0,sothat\underset{(n\in \mathbb{P})\to \mathrm{\infty }}{\lim }{a}_n\ne 0\Rightarrow \mathrm{\Sigma }{a}_{n}does$$$$notconverge.$$$$$$$$Supposethatn\in \mathbb{P}-\left\{2\right\}and{z}^n=e^{2\pi ir}with$$$$r=0,1,2,\dots ,n-1.Then,z=e^{\frac{2\pi ir}{n}}.$$$$n\in \mathbb{P}-\left\{2\right\}\Rightarrow \frac{r}{n}neverreducesnorvanishes$$$$forallr\ne 0.Ifr=0weobtainthenon-primitive$$$$rootz=1.Otherwise,allarguments$$$$\frac{2\pi r}{n}areunique.Inaddingallofthe$$$$primitiverootshowever,theimaginary$$$$partsvanishforthereareaneven$$$$numberofroots(n-1\because n\in {\mathbb{O}}^{+})andhalfofthese$$$$rootsarecomplexconjugatesoftheremaining$$$$roots.$$$$\therefore a_n=2\underset{r=1}{\sum ^{\frac{n-1}{2}}}cos\frac{2\pi r}{n}(n\in \mathbb{P}-\left\{2\right\})$$$$\Rightarrow \underset{(n\in \mathbb{P}-\left\{2\right\})\to \mathrm{\infty }}{\lim }{a}_n\ne 0$$$$Ifn=2,a_2=-1\ne 0.So\underset{(n\in \mathbb{P})\to \mathrm{\infty }}{\lim }{a}_n\ne 0.$$$$Thereexistinfinitelymanyprimes$$$$with\mathbb{P}\subset \mathbb{N},so{a}_{n}alwaysexistsfor$$$$n\in \mathbb{P}\subset \mathbb{N}.$$$$Calculationsprojectthat{a}_n=-1$$$$foralln\in \mathbb{P}.Ithasbeenproventhat$$$$a_n=-1\mathrm{\forall }n\in \mathbb{P}indeedinthecomments$$$$below.$$$$$$$$$$$$$$$$$$$$$$$$$$

Commented by prakash jain last updated on 18/Dec/15

$$\mathrm{Thanks}\mathrm{for}\mathrm{detailed}\mathrm{answer}.$$

Commented by Yozzii last updated on 18/Dec/15

$$S\left(n\right)=\underset{r=1}{\sum ^n}cosr\theta$$$$S\left(n\right)sin0.5\theta =\underset{r=1}{\sum ^n}cosr\theta sin0.5\theta (\theta \ne 2n\pi )$$$$$$$$2cosr\theta sin0.5\theta =sin(r+0.5)\theta -sin(r-0.5)\theta$$$$\therefore S\left(n\right)sin0.5\theta =\frac{1}{2}\{sin1.5\theta -sin0.5\theta$$$$+sin2.5\theta -sin1.5\theta +sin3.5\theta -sin2.5\theta$$$$+\dots +sin(n-2.5)\theta -sin(n-3.5)\theta$$$$+sin(n-1.5)\theta -sin(n-2.5)\theta$$$$+sin(n-0.5)\theta -sin(n-1.5)\theta$$$$+sin(n+0.5)\theta -sin(n-0.5)\theta \}$$$$S\left(n\right)=\frac{sin(n+0.5)\theta -sin0.5\theta }{2sin0.5\theta }$$$$S\left(n\right)=\frac{2cos\frac{(n+0.5)\theta +0.5\theta }{2}sin\frac{(n+0.5)\theta -0.5\theta }{2}}{2sin0.5\theta }$$$$S\left(n\right)=\frac{cos\frac{n+1}{2}\theta sin\frac{n\theta }{2}}{sin0.5\theta }$$$$n\in \mathbb{P}-\left\{2\right\}\Rightarrow \frac{n-1}{2}\in {\mathbb{Z}}^{+}$$$$so\frac{n-1}{2}canbesafelysubstitutedforninS\left(n\right).$$$$$$$$\therefore S\left(\frac{n-1}{2}\right)=\frac{cos\frac{\frac{n}{2}-\frac{1}{2}+1}{2}\theta sin\frac{n-1}{4}\theta }{sin0.5\theta }$$$$S\left(\frac{n-1}{2}\right)=\frac{cos\frac{n+1}{4}\theta sin\frac{n-1}{4}\theta }{2cos\frac{\theta }{4}sin\frac{\theta }{4}}$$$$$$$$Nowlet\theta =\frac{2\pi }{n}.$$$$S(0.5(n-1))=\frac{cos\frac{n+1}{2n}\pi sin\frac{n-1}{2n}\pi }{2cos\frac{\pi }{2n}sin\frac{\pi }{2n}}$$$$S\left(\frac{1}{2}(n-1)\right)=\frac{cos(\frac{\pi }{2}+\frac{\pi }{2n})sin(\frac{\pi }{2}-\frac{\pi }{2n})}{2cos\frac{\pi }{2n}sin\frac{\pi }{2n}}$$$$cos(0.5\pi +\alpha )=0-sin0.5\pi sin\alpha$$$$andsin(0.5\pi -\alpha )=cos\alpha$$$$\Rightarrow S\left(\frac{1}{2}(n-1)\right)=-0.5$$$$\Rightarrow a_n=2\times (-0.5)=-1\mathrm{\forall }n\in \mathbb{P}-\left\{2\right\}.$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$o$$

Commented by prakash jain last updated on 18/Dec/15

$$p>2prime-\left(soodd\right)$$$$x^p-1=0$$$$1+x+x^2+..+x^{p-1}=0\left(1\right)$$$$\mathrm{All}\mathrm{roots}\mathrm{of}\left(1\right)\mathrm{are}\mathrm{primitive}$$$$(x-r_1)(x-r_2)(x-r_3)\dots (x-r_{p-1})=0$$$$x^{p-1}-(r_1+r_2+..+r_{p-1})x^{p-2}+\dots .+{(-1)}^{p-1}\underset{i=1}{\prod ^{p-1}}{r}_i=0$$$$so$$$$r_1+r_2+..+r_{p-1}=-1$$$$\mathrm{Same}\mathrm{as}\mathrm{what}\mathrm{you}\mathrm{stated}\mathrm{that}\mathrm{sum}\mathrm{of}\mathrm{primitive}$$$$\mathrm{roots}=-1\mathrm{for}n\in \mathbb{P}.$$

Commented by Yozzii last updated on 18/Dec/15

$$Ahh.Nicely.$$

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