If a_n is sum of the all primitive n^(th) root of unity. Does Σ_(n=1) ^∞ a_n converge? An n^(th) root of unity, say z, is primitive if it is not k^(th) root of unity where k<n. or z^n =1 and z^k ≠1 (k=1,2,3,..,n−1) a_1 =1 a_2 =−1 (2^(nd) root of unity are 1 and −1, 1 is not primitive)


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

$If a_{n} is sum of the all primitive n^{t h} root of$ $unity .$ $Does \underset{n = 1}{\sum^{\infty}} a_{n} converge ?$ $An n^{t h} root of unity, say z, is primitive if it is not$ $k^{t h} root of unity where k < n .$ $o r z^{n} = 1 and z^{k} \neq 1 (k = 1, 2, 3, . ., n - 1)$ $a_{1} = 1$ $a_{2} = - 1 (2^{n d} root of unity are 1 and - 1, 1 is not$ $primitive)$
	Answered by Yozzii last updated on 18/Dec/15
	$Let p(n) be the primitive roots of the equation z^n =1 and a_n =Σp(n). n=1⇒z=1⇒p(1)=1⇒a_1 =1. n=2⇒z^2 =1⇒z=±1⇒p(2)=−1⇒a_2 =−1. n=3⇒z^3 =1=e^(2πir) (−1≤r≤1) ⇒z=e^((2πir)/3) ⇒p(3)=e^(±((2πi)/3)) ⇒a_3 =2cos((2π)/3)=−1 n=4⇒z^4 =1⇒z=±1,±i⇒p(4)=±i⇒a_4 =0 n=5⇒z^5 =e^(2πir) (−2≤r≤2) ⇒z=e^((2πir)/5) =1,e^(±((2πi)/5)) ,e^(±((4πi)/5)) ⇒p(5)=e^(±((2πi)/5)) ,e^(±((4πi)/5)) ⇒a_5 =2(cos((2π)/5)+cos((4π)/5)) a_5 =2{2cos(((4π+2π)/5)/2)cos(((4π−2π)/5)/2)} a_5 =4cos((3π)/5)cos(π/5)=−1 n=6⇒z^6 =e^(2πir) (−2≤r≤3) z=e^((πir)/3) ⇒z=±1,e^(±((πi)/3)) ,e^(±((2πi)/3)) ⇒p(6)=e^(±((πi)/3)) ⇒a_6 =2cos(π/3)=1 n=7⇒z^7 =e^(2πir) (−3≤r≤3) ⇒z=e^((2πir)/7) ⇒z=1,e^((±2πi)/7) ,e^((±4πi)/7) ,e^((±6πi)/7) ⇒a_7 =2(cos((2π)/7)+cos((4π)/7)+cos((6π)/7))=−1 n=8⇒z^8 =e^(2πir) ⇒z=e^((πir)/4) (−3≤r≤4) ⇒z=±1,e^(±((3πi)/4)) ,±i,e^(±((πi)/4)) ⇒a_8 =2(cos((3π)/4)+cos(π/4))=0. n=9⇒z=e^((2πir)/9) (−4≤r≤4) ⇒z=1,e^(±((2πi)/9)) ,e^(±((4πi)/9)) ,e^(±((2πi)/3)) ,e^(±((8πi)/9)) ⇒a_9 =2(cos((2π)/9)+cos((4π)/9)+cos((8π)/9))=0 n=10⇒z=e^((πir)/5) (−4≤r≤5) ⇒z=±1,e^(±((πi)/5)) ,e^(±((2πi)/5)) ,e^(±((3πi)/5)) ,e^(±((4πi)/5)) ⇒a_(10) =2(cos(π/5)+cos((3π)/5))=1 I conject that for n being prime, a_n ≠0, so that lim_((n∈P)→∞) a_n ≠0⇒Σa_n does not converge. Suppose that n∈P−{2} and z^n =e^(2πir) with r=0,1,2,...,n−1. Then, z=e^((2πir)/n) . n∈P−{2}⇒ (r/n) never reduces nor vanishes for all r≠0. If r=0 we obtain the non−primitive root z=1. Otherwise, all arguments ((2πr)/n) are unique. In adding all of the primitive roots however, the imaginary parts vanish for there are an even number of roots (n−1 ∵ n∈O^+ ) and half of these roots are complex conjugates of the remaining roots. ∴ a_n =2Σ_(r=1) ^((n−1)/2) cos((2πr)/n) (n∈P−{2}) ⇒lim_((n∈P−{2})→∞) a_n ≠0 If n=2, a_2 =−1≠0. So lim_((n∈P)→∞) a_n ≠0. There exist infinitely many primes with P⊂N, so a_n always exists for n∈P⊂N. Calculations project that a_n =−1 for all n∈P. It has been proven that a_n =−1 ∀n∈P indeed in the comments below.$
	$L e t p (n) b e t h e p r i m i t i v e r o o t s o f t h e$ $e q u a t i o n z^{n} = 1 a n d a_{n} = Σ p (n) .$ $n = 1 \Rightarrow z = 1 \Rightarrow p (1) = 1 \Rightarrow a_{1} = 1 .$ $n = 2 \Rightarrow z^{2} = 1 \Rightarrow z = \pm 1 \Rightarrow p (2) = - 1 \Rightarrow a_{2} = - 1 .$ $n = 3 \Rightarrow z^{3} = 1 = e^{2 π i r} (- 1 ⩽ r ⩽ 1)$ $\Rightarrow z = e^{\frac{2 π i r}{3}} \Rightarrow p (3) = e^{\pm \frac{2 π i}{3}} \Rightarrow a_{3} = 2 c o s \frac{2 π}{3} = - 1$ $n = 4 \Rightarrow z^{4} = 1 \Rightarrow z = \pm 1, \pm i \Rightarrow p (4) = \pm i \Rightarrow a_{4} = 0$ $n = 5 \Rightarrow z^{5} = e^{2 π i r} (- 2 ⩽ r ⩽ 2)$ $\Rightarrow z = e^{\frac{2 π i r}{5}} = 1, e^{\pm \frac{2 π i}{5}}, e^{\pm \frac{4 π i}{5}} \Rightarrow p (5) = e^{\pm \frac{2 π i}{5}}, e^{\pm \frac{4 π i}{5}}$ $\Rightarrow a_{5} = 2 (c o s \frac{2 π}{5} + c o s \frac{4 π}{5})$ $a_{5} = 2 {2 c o s \frac{\frac{4 π + 2 π}{5}}{2} c o s \frac{\frac{4 π - 2 π}{5}}{2}}$ $a_{5} = 4 c o s \frac{3 π}{5} c o s \frac{π}{5} = - 1$ $n = 6 \Rightarrow z^{6} = e^{2 π i r} (- 2 ⩽ r ⩽ 3)$ $z = e^{\frac{π i r}{3}} \Rightarrow z = \pm 1, e^{\pm \frac{π i}{3}}, e^{\pm \frac{2 π i}{3}} \Rightarrow p (6) = e^{\pm \frac{π i}{3}}$ $\Rightarrow a_{6} = 2 c o s \frac{π}{3} = 1$ $n = 7 \Rightarrow z^{7} = e^{2 π i r} (- 3 ⩽ r ⩽ 3)$ $\Rightarrow z = e^{\frac{2 π i r}{7}} \Rightarrow z = 1, e^{\frac{\pm 2 π i}{7}}, e^{\frac{\pm 4 π i}{7}}, e^{\frac{\pm 6 π i}{7}}$ $\Rightarrow a_{7} = 2 (c o s \frac{2 π}{7} + c o s \frac{4 π}{7} + c o s \frac{6 π}{7}) = - 1$ $n = 8 \Rightarrow z^{8} = e^{2 π i r} \Rightarrow z = e^{\frac{π i r}{4}} (- 3 ⩽ r ⩽ 4)$ $\Rightarrow z = \pm 1, e^{\pm \frac{3 π i}{4}}, \pm i, e^{\pm \frac{π i}{4}}$ $\Rightarrow a_{8} = 2 (c o s \frac{3 π}{4} + c o s \frac{π}{4}) = 0 .$ $n = 9 \Rightarrow z = e^{\frac{2 π i r}{9}} (- 4 ⩽ r ⩽ 4)$ $\Rightarrow z = 1, e^{\pm \frac{2 π i}{9}}, e^{\pm \frac{4 π i}{9}}, e^{\pm \frac{2 π i}{3}}, e^{\pm \frac{8 π i}{9}}$ $\Rightarrow a_{9} = 2 (c o s \frac{2 π}{9} + c o s \frac{4 π}{9} + c o s \frac{8 π}{9}) = 0$ $n = 10 \Rightarrow z = e^{\frac{π i r}{5}} (- 4 ⩽ r ⩽ 5)$ $\Rightarrow z = \pm 1, e^{\pm \frac{π i}{5}}, e^{\pm \frac{2 π i}{5}}, e^{\pm \frac{3 π i}{5}}, e^{\pm \frac{4 π i}{5}}$ $\Rightarrow a_{10} = 2 (c o s \frac{π}{5} + c o s \frac{3 π}{5}) = 1$ $I c o n j e c t t h a t f o r n b e i n g p r i m e,$ $a_{n} \neq 0, s o t h a t \lim_{(n \in P) \to \infty} a_{n} \neq 0 \Rightarrow Σ a_{n} d o e s$ $n o t c o n v e r g e .$ $S u p p o s e t h a t n \in P - {2} a n d z^{n} = e^{2 π i r} w i t h$ $r = 0, 1, 2, . . ., n - 1 . T h e n, z = e^{\frac{2 π i r}{n}} .$ $n \in P - {2} \Rightarrow \frac{r}{n} n e v e r r e d u c e s n o r v a n i s h e s$ $f o r a l l r \neq 0 . I f r = 0 w e o b t a i n t h e n o n - p r i m i t i v e$ $r o o t z = 1 . O t h e r w i s e, a l l a r g u m e n t s$ $\frac{2 π r}{n} a r e u n i q u e . I n a d d i n g a l l o f t h e$ $p r i m i t i v e r o o t s h o w e v e r, t h e i m a g i n a r y$ $p a r t s v a n i s h f o r t h e r e a r e a n e v e n$ $n u m b e r o f r o o t s (n - 1 ∵ n \in O^{+}) a n d h a l f o f t h e s e$ $r o o t s a r e c o m p l e x c o n j u g a t e s o f t h e r e m a i n i n g$ $r o o t s .$ $∴ a_{n} = 2 \underset{r = 1}{\sum^{\frac{n - 1}{2}}} c o s \frac{2 π r}{n} (n \in P - {2})$ $\Rightarrow \lim_{(n \in P - {2}) \to \infty} a_{n} \neq 0$ $I f n = 2, a_{2} = - 1 \neq 0 . S o \lim_{(n \in P) \to \infty} a_{n} \neq 0 .$ $T h e r e e x i s t i n f i n i t e l y m a n y p r i m e s$ $w i t h P \subset N, s o a_{n} a l w a y s e x i s t s f o r$ $n \in P \subset N .$ $C a l c u l a t i o n s p r o j e c t t h a t a_{n} = - 1$ $f o r a l l n \in P . I t h a s b e e n p r o v e n t h a t$ $a_{n} = - 1 \forall n \in P i n d e e d i n t h e c o m m e n t s$ $b e l o w .$
	Commented byprakash jain last updated on 18/Dec/15

	$Thanks for detailed answer .$
	Commented byYozzii last updated on 18/Dec/15
	$S(n)=Σ_(r=1) ^n cosrθ S(n)sin0.5θ=Σ_(r=1) ^n cosrθsin0.5θ (θ≠2nπ) 2cosrθsin0.5θ=sin(r+0.5)θ−sin(r−0.5)θ ∴S(n)sin0.5θ=(1/2){sin1.5θ−sin0.5θ +sin2.5θ−sin1.5θ+sin3.5θ−sin2.5θ +...+sin(n−2.5)θ−sin(n−3.5)θ +sin(n−1.5)θ−sin(n−2.5)θ +sin(n−0.5)θ−sin(n−1.5)θ +sin(n+0.5)θ−sin(n−0.5)θ} S(n)=((sin(n+0.5)θ−sin0.5θ)/(2sin0.5θ)) S(n)=((2cos(((n+0.5)θ+0.5θ)/2)sin(((n+0.5)θ−0.5θ)/2))/(2sin0.5θ)) S(n)=((cos((n+1)/2)θsin((nθ)/2))/(sin0.5θ)) n∈P−{2}⇒ ((n−1)/2)∈Z^+ so ((n−1)/2) can be safely substituted for n in S(n). ∴ S(((n−1)/2))=((cos(((n/2)−(1/2)+1)/2)θsin((n−1)/4)θ)/(sin0.5θ)) S(((n−1)/2))=((cos((n+1)/4)θsin((n−1)/4)θ)/(2cos(θ/4)sin(θ/4))) Now let θ=((2π)/n). S(0.5(n−1))=((cos((n+1)/(2n))πsin((n−1)/(2n))π)/(2cos(π/(2n))sin(π/(2n)))) S((1/2)(n−1))=((cos((π/2)+(π/(2n)))sin((π/2)−(π/(2n))))/(2cos(π/(2n))sin(π/(2n)))) cos(0.5π+α)=0−sin0.5πsinα and sin(0.5π−α)=cosα ⇒S((1/2)(n−1))=−0.5 ⇒a_n =2×(−0.5)=−1 ∀n∈P−{2}. o$
	$S (n) = \underset{r = 1}{\sum^{n}} c o s r θ$ $S (n) s i n 0.5 θ = \underset{r = 1}{\sum^{n}} c o s r θ s i n 0.5 θ (θ \neq 2 n π)$ $2 c o s r θ s i n 0.5 θ = s i n (r + 0.5) θ - s i n (r - 0.5) θ$ $∴ S (n) s i n 0.5 θ = \frac{1}{2} {s i n 1.5 θ - s i n 0.5 θ$ $+ s i n 2.5 θ - s i n 1.5 θ + s i n 3.5 θ - s i n 2.5 θ$ $+ . . . + s i n (n - 2.5) θ - s i n (n - 3.5) θ$ $+ s i n (n - 1.5) θ - s i n (n - 2.5) θ$ $+ s i n (n - 0.5) θ - s i n (n - 1.5) θ$ $+ s i n (n + 0.5) θ - s i n (n - 0.5) θ}$ $S (n) = \frac{s i n (n + 0.5) θ - s i n 0.5 θ}{2 s i n 0.5 θ}$ $S (n) = \frac{2 c o s \frac{(n + 0.5) θ + 0.5 θ}{2} s i n \frac{(n + 0.5) θ - 0.5 θ}{2}}{2 s i n 0.5 θ}$ $S (n) = \frac{c o s \frac{n + 1}{2} θ s i n \frac{n θ}{2}}{s i n 0.5 θ}$ $n \in P - {2} \Rightarrow \frac{n - 1}{2} \in Z^{+}$ $s o \frac{n - 1}{2} c a n b e s a f e l y s u b s t i t u t e d f o r n i n S (n) .$ $∴ S (\frac{n - 1}{2}) = \frac{c o s \frac{\frac{n}{2} - \frac{1}{2} + 1}{2} θ s i n \frac{n - 1}{4} θ}{s i n 0.5 θ}$ $S (\frac{n - 1}{2}) = \frac{c o s \frac{n + 1}{4} θ s i n \frac{n - 1}{4} θ}{2 c o s \frac{θ}{4} s i n \frac{θ}{4}}$ $N o w l e t θ = \frac{2 π}{n} .$ $S (0.5 (n - 1)) = \frac{c o s \frac{n + 1}{2 n} π s i n \frac{n - 1}{2 n} π}{2 c o s \frac{π}{2 n} s i n \frac{π}{2 n}}$ $S (\frac{1}{2} (n - 1)) = \frac{c o s (\frac{π}{2} + \frac{π}{2 n}) s i n (\frac{π}{2} - \frac{π}{2 n})}{2 c o s \frac{π}{2 n} s i n \frac{π}{2 n}}$ $c o s (0.5 π + α) = 0 - s i n 0.5 π s i n α$ $a n d s i n (0.5 π - α) = c o s α$ $\Rightarrow S (\frac{1}{2} (n - 1)) = - 0.5$ $\Rightarrow a_{n} = 2 \times (- 0.5) = - 1 \forall n \in P - {2} .$ $o$
	Commented byprakash jain last updated on 18/Dec/15

	$p > 2 p r i m e - (s o o d d)$ $x^{p} - 1 = 0$ $1 + x + x^{2} + . . + x^{p - 1} = 0 (1)$ $All roots of (1) are primitive$ $(x - r_{1}) (x - r_{2}) (x - r_{3}) . . . (x - r_{p - 1}) = 0$ $x^{p - 1} - (r_{1} + r_{2} + . . + r_{p - 1}) x^{p - 2} + . . . . + {(- 1)}^{p - 1} \underset{i = 1}{\prod^{p - 1}} r_{i} = 0$ $s o$ $r_{1} + r_{2} + . . + r_{p - 1} = - 1$ $Same as what you stated that sum of primitive$ $roots = - 1 for n \in P .$
	Commented byYozzii last updated on 18/Dec/15

	$A h h . N i c e l y .$
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