# On the Uniform Convergence of Sine Series with Square Root

Authors: Sergiusz Kęska
Journal: Journal of Function Spaces
 (2019)

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## Abstract

Chaundy and Jolliffe proved that if{ck}k=1∞ is a nonincreasing real sequence with limk→∞ck=0, then the series ∑k=1∞‍cksin⁡kx converges uniformly if and only if kck→0. The purpose of this paper is to show that kck→0 is a necessary and sufficient condition for the uniform convergence of series ∑k=1∞‍cksin⁡kθ in θ∈[0,π]. However for ∑k=1∞‍cksin⁡k2θ it is not true in θ∈[0,π].


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## Body

## 1. Introduction

Chaundy and Jolliffe [1] proved the following.Theorem 1.
If{ck}k=1∞⊂R+ is decreasing to zero, then ∑k=1∞‍cksin⁡kx converges uniformly in x if and only if kck→0 as k→∞.Theorem1 has had numerous generalizations.Leindler [2] verified that in Theorem 1 the monotonicity assumption cn≥cn+1 can be replaced by c∈RBVS, i.e., if the conditions cn→0 and ∑k=n∞‍ck-ck+1≤Kcn hold for all n with constant K=K(c) which depends only upon c.The next theorem was indicated in [3].Theorem 2.
If{ck} belongs to the class MVBVS, i.e., if there exist constants C and λ≥2, depending only on the sequence {ck} such that
∑ k = n 2 n ‍ c k - c k + 1 ≤ C / n ∑ k = ⌊ λ - 1 n ⌋ ⌊ λ n ⌋ ‍ c k for all n≥λ, then series ∑k=1∞‍cksin⁡kx converges uniformly in x if and only if limk→∞kck=0.Móricz [4] proves the following theorem.Theorem 3.
Assumef:R+→[0,∞) with property xf(x)∈Lloc1(R+). If f(x) is nonincreasing on R+, then integral ∫0∞‍fxsin⁡txdx, t∈R+, converges uniformly in t if and only if xf(x)→0 as x→∞.A result due toZ˘ak and S˘neider [5] holds for double sine series. ∑j=1∞‍∑k=1∞‍cjksin⁡jxsin⁡ky is regularly convergent in case of a fixed (x,y) if the rectangular sums ∑j=1m‍∑k=1n‍cjksin⁡jxsin⁡ky converge to a finite number as m and n independently tend to infinity; moreover the row and column series ∑j=1∞‍cjnsin⁡jxsin⁡ny, n=1,2,…, and ∑k=1∞‍cmksin⁡mxsin⁡ky, m=1,2,…, are convergent.Theorem 4.
If{cjk}j,k=1∞⊂R+ is a monotonically decreasing double sequence, i.e., sequence of real numbers such that, for j,k=1,2,…,
c j k - c j + 1 , k ≥ 0, cjk-cj,k+1≥0, and cjk-cj,k+1-cj+1,k+cj+1,k+1≥0, then ∑j=1∞‍∑k=1∞‍cjksin⁡jxsin⁡ky is uniformly regularly convergent in (x,y) if and only if jkcjk→0 as j+k→∞.Theorem4 was generalized by Kórus [6]. He has defined new classes of double sequences (SBVDS1) to obtain those generalizations.Duzinkiewicz and Szal [7] introduce a new class of double sequence called DGM(α,β,γ,r), which is a generalization of the class considered by Kórus, and they obtain sufficient and necessary conditions for uniform convergence of double sine series.A series(1)∑ckeiλkθwas motivation for the generalization of the Theorem 1. Such series were studied by Paley and Wiener who called them nonharmonic Fourier series. They proved the following [8].Theorem 5.
Ifλk-k≤D<1/π2 for -∞<k<∞, then the sequence {eiλkθ} is closed in L2(-π,π) and possesses a unique biorthogonal set {hk(θ)}, such that the series
∑ k = - ∞ ∞ ‍ { e i λ k θ / 2 π ∫ - π π ‍ f ( t ) e - i k t d t - e i λ k θ ∫ - π π ‍ f ( t ) h k ( t ) d t } converges uniformly to zero over interval (-π+δ,π-δ) for any positive δ, and over any such interval the summability properties of
∑ k = - ∞ ∞ ‍ e i λ k θ ∫ - π π ‍ f ( t ) h k ( t ) d t are uniformly the same as those of the Fourier series of f(θ).We will consider a special case of the series (1) for λk=k and λk=k2, k≥1, which does not meet the assumptions of the above theorem.


## 2. Main Results

Theorem 6.
If{ck}k=1∞⊂R+ is nonincreasing, then the series ∑k=1∞‍cksin⁡kθ converges uniformly in θ∈[0,π] if and only if limk→∞kck=0.Proof (necessary condition).
Suppose that a series∑k=1∞‍cksin⁡kθ converges uniformly on [0,π]. Let θ=π/α. We consider α=n for n=4r and r∈N={1,2,3,…}: ∑k=5n/49n/4‍cksin⁡kπ/α=∑k=5n/49n/4‍cksin⁡kπ/n=∑k=5n/49n/4‍cksin⁡π-k/nπ≤∑k=5n/49n/4‍2ck(1-k/n)=2(c5n/4(1-5/2)+…+c9n/4(1-3/2))≤2(1-5/2)(c5n/4+…+c9n/4))≤(1-5/2)nc9n/4=4/9(1-5/2)9rc9r.
Hence,(2)∀r∈N∑k=5r9rcksin⁡π2k4r≥495/2-19rc9r.After considering that ∑k=1∞‍cksin⁡kθ converges uniformly we obtain
∑ k = 5 r k = 9 r ‍ c k sin ⁡ π 2 k / 4 r < ϵ for sufficiently large r. Thus, in view of inequality (2), we obtain 4/9(5/2-1)9rc9r<ϵ for sufficiently large r, so(3)limr→∞9rc9r=0.After considering that the sequence {ck} is nonincreasing we have(4)9rc9r≤9rc9r-1≤…≤9rc9r-8≤9rc9r-1=9r-1c9r-1+9c9r-1.Thus(5)∀s=0,1,…,8limr→∞9rc9r-s=0by (3). In view of (5), we obtain ∀s=0,1,…,8limr→∞9r-sc9r-s=limr→∞9rc9r-s-limr→∞sc9r-s=0 and limm→∞mcm=0.Proof (sufficient condition).
Letθ=π/α.
Case 1:(6)α≥p≥n.After considering k/α≤p/p=1 and sin⁡πk/α≤πk/α we obtain(7)0≤∑k=npcksin⁡πkα≤πα∑k=npckk=παncnn+…+pcpp≤παsupk≥nkck1n+…+1p≤πsupk≥nkck1p∑k=1p1k.This follows from (6). Note that the following condition is fulfilled:(8)1+12+…+1p<1p+1pp+1-pforp≥1. In view of (8) the following inequality is satisfied for bp=1/p∑k=1p‍1/k:(9)bp+1bp=1p+1p+pp+111+1/2+…+1/p>1p+1p+pp+1p+1-pp/p+1=1.Thus the sequence bp is increasing with respect to p and limp→∞bp=2.(10)∀n≤p1p∑k=np1k≤limp→∞1p∑k=1p1k=2. This follows from (9). Finally for n≤p and α≥p,(11)0≤∑k=npcksin⁡πkα≤2πsupk≥nkck.This follows from (7), (10).
To prove the caseα≤p we first observe the following.
Lemma 2.2.Letα≥1andm∈N. Letdenote the floor function, i.e.,x=z⇔z∈Zandx-1<z≤x. Then(12)∀s=1,2,3,…,α2m+34-1α2m+32≤α2m+12-s+α2m+12+s≤2αm+1(13)∀s=1,2,3,…,α2m+34-1-α≤α2m+12-s-α2m+12+s≤0(14)∀s=α2m+34+2,…,α22m+1-1α2m+3≥α2m+12+1-s+α2m+12+2α2+1+s≥2αm+1(15)∀s=α2m+34+2,…,α22m+1-1-2α≤α2m+12+1-s-α2m+12+2α2+1+s≤-α.
The proof of (12).
Note that fors=0,1,2,…,α2(m+3/4) the following condition is fulfilled:(16)α2m+12-s+α2m+12+s≤2αm+1,which follows from the relationship: (17)α2m+12-s+α2m+12+s+2α4m+14-s2≤4α2m+12andα4m+14-s2≤α4m+14.For s=1,2,…,⌊α2(m+3/4)⌋-1,(18)α2m+12-s+α2m+12+s≤α2m+12-s+α2m+12+s≤2αm+1.This follows from (16).
The proof of (14).
Note that fors=α2(m+3/4),…,α2(2m+1) the following condition is fulfilled:(19)α2m+12-s+α2m+12+2α2+s≥2αm+1,which follows from the relationship:(20)2α2+2α2m+12+2α4m+14-s2-2α2s+2α4m+12≥4α2m+12(21)andα4m+14-s2-sα2s+2α4m+12≥α4m+14-2α4m+12+α4for s∈[0,α2(2m+1)].
Fors=⌊α2(m+3/4)⌋+2,…,⌊α2(2m+1)⌋-1,(22)α2m+12+1-s+α2m+12+2α2+1+s≥α2m+12-s+α2m+12+2α2+s≥2αm+1.This follows from (19). The proof of the rest of Lemma 2.2 is obvious.
Case 2:(23)1≤α≤n≤p.Case 2’:(24)pn<7.Therefore,(25)∑k=npcksin⁡πkα≤p-n+1cn≤49ncn≤49supk≥nkck.Case 2":(26)pn≥7.In view of (23), for all α there are(27)anoddnumbermn′α≥1that:mn′α-2<nα≤mn′α,(28)anevennumbermn′′α≥2that:mn′′α-2<nα≤mn′′α.Note that(29)∑k=nα2mn′2αcksin⁡πkα≤∑k=nα2mn′2αck≤α2mn′2α-n+1cn=α2mn′α-22+4α2mn′α-4α2-n+1cn≤n+4α2mn′α-1-n+1cn=4α2mn′α-2+4α2+1cn≤4αn+4α2+1cn≤9ncn≤9supk≥nkck,which follows from (23) and (27). The proof of(30)∑k=nα2mn′′2αcksin⁡πkα≤9supk≥nkckis similar. This follows from (23) and (28). Moreover for all α there are(31)anoddnumbermp′α≥1that:mp′α≤pα<mp′α+2,(32)anevennumbermp′′α≥2that:mp′′α≤pα<mp′′α+2.Note that(33)∑k=α2mp′2α+1pcksin⁡πkα≤p-α2mp′αcα2mp′2α+1≤p-α2mp′α+1cα2mp′2α+1≤p-α2mp′2α+1α2mp′2α+1α2mp′2α+1cα2mp′2α+1.Let us observe that p/α-n/α=n/α(p/n-1)≥6. This follows from (23) and (26). Hence,(34)pα-4≥nα+2.Therefore,(35)mp′α-2>pα-4≥nα+2>mn′α.This follows from (27), (31), and (34). The proof of(36)mp′′α-2>mn′′αis similar. This follows from (28), (32), and (34).
Furthermoreα2mp′2α+1≥n. This follows from (27), (35). Thus,(37)α2mp′2α+1cα2mp′2α+1≤supk≥nkck.Note that(38)p-α2mp′2α+1α2mp′2α+1≤pα2mp′2α-1+2α2mp′2α≤pαmp′α+2-2α2+1≤pp-2α2+1=11-2α/p2+1≤11-2n/p2+1≤11-2/72+1<3,which follows from the relationship α(mp′(α)+2)>p≥7n≥7α,α≥1 and 2α≤2n≤2/7p (this follows from (23), (26), and (31)).
Therefore, forp/n≥7,(39)∑k=α2mp′2α+1pcksin⁡πkα<3supk≥nkck. This follows from (33), (37), and (38). In analogy with (33), (37), and (38) we have(40)∑k=α2mp′′2α+1pcksin⁡πkα≤p-α2mp′′2α+1α2mp′′2α+1α2mp′′2α+1cα2mp′′2α+1;(41)α2mp′′2α+1cα2mp′′2α+1≤supk≥nkck,this follows from (28) and (36);(42)p-α2mp′′2α+1α2mp′′2α+1<3,which follows from the relationship
α ( m p ′ ′ ( α ) + 2 ) > p ≥ 7 n ≥ 7 α , α ≥ 1 and 2α≤2n≤2/7p (this follows from (23), (26), and (32)). Therefore, for p/n≥7,(43)∑k=α2mp′′2α+1pcksin⁡πkα<3supk≥nkck. This follows from (40), (41), and (42). Denote by S the sum ∑k=np‍cksin⁡πk/α. Let us observe that(44)S=∑k=nα2mn′2αcksin⁡πkα+∑k=α2mp′2αpcksin⁡πkα+S′,whereS′=∑m=mn′αm∈Oddnumbersmp′α-2∑z=α2m2+1α2m+1/22+∑z=α2m+1/22+1α2m+12+∑z=α2m+12+1α2m+3/22+∑z=α2m+3/22+1α2m+22czsin⁡πzα.Then,(45)S≤∑k=nα2mn′2αcksin⁡πkα+∑k=α2mp′2αpcksin⁡πkα+S′≤12supk≥nkck+S′.This follows from (29) and (39). Note that for any odd number m and ∀α≥1 we have(46)cα2m+12∑z=α2m2+1α2m+1/22sin⁡πzα+∑z=α2m+1/22+1α2m+12sin⁡πzα≥∑z=α2m2+1α2m+1/22czsin⁡πzα+∑z=α2m+1/22+1α2m+12czsin⁡πzα,which follows from the relationship(47)sin⁡πα2m2+1α≤0,…,sin⁡πα2m+12α≤0.On the other hand, for any odd number m and ∀α≥1, we have(48)cα2m+12∑z=α2m+12+1α2m+3/22sin⁡πzα+∑z=α2m+3/22+1α2m+22sin⁡πzα≥∑z=α2m+12+1α2m+3/22czsin⁡πzα+∑z=α2m+3/22+1α2m+22czsin⁡πzα,which follows from the relationship(49)sin⁡πα2m+12+1α≥0,…,sin⁡πα2m+22α≥0.Thus(50)S′≤∑m=mn′αm∈Oddnumbersmp′α-2cα2m+12∑z=α2m2+1α2m+22sin⁡πzα.This follows from (46) and (48). Note that ∀m∈N we have(51)∑s=1α2m+3/4-1sin⁡πα2m+12-sα+sin⁡πα2m+12+sα=2∑s=1α2m+3/4-1sin⁡πα2m+12-s+α2m+12+s2αcos⁡πα2m+12-s-α2m+12+s2α.We see that ∀s=1,2,…,⌊α2(m+3/4)⌋-1, m∈N and α≥1,(52)m+34π≤π2αα2m+12-s+α2m+12+s≤m+1π.This follows from (12). Thus ∀s=1,2,…,⌊α2(m+3/4)⌋-1, α≥1, and for any odd number m we have(53)-1≤sin⁡πα2m+12-s+α2m+12+s2α≤0.Moreover ∀s=1,2,…,⌊α2(m+3/4)⌋-1, m∈N and α≥1, we get(54)-π2≤π2αα2m+12-s-α2m+12+s≤0.This follows from (13). Thus ∀s=1,2,…,⌊α2(m+3/4)⌋-1, α≥1, and for any odd number m we have(55)0≤cos⁡πα2m+12-s-α2m+12+s2α≤1.In view of (51), (53), and (55), for any odd number m, the following inequality is satisfied:(56)∑s=1α2m+3/4-1sin⁡πα2m+12-sα+sin⁡πα2m+12+sα≤0.Note that ∀m∈N(57)∑s=α2m+3/4+2α22m+1-1sin⁡πα2m+12+1-sα+sin⁡πα2m+12+2α2+1+sα=2∑s=α2m+3/4+2α22m+1-1sin⁡πα2m+12+1-s+α2m+12+2α2+1+s2αcos⁡πα2m+12+1-s-α2m+12+2α2+1+s2α.We see that ∀s=⌊α2(m+3/4)⌋+2,…,⌊α2(2m+1)⌋-1, m∈N and α≥1,(58)m+1π≤π2αα2m+12+1-s+α2m+12+2α2+1+s≤m+32π.This follows from (14). Thus ∀s=⌊α2(m+3/4)⌋+2,…,⌊α2(2m+1)⌋-1, α≥1, and for any odd number m we have(59)0≤sin⁡πα2m+12+1-s2α+πα2m+12+2α2+1+s2α≤1.We see that ∀s=⌊α2(m+3/4)⌋+2,…,⌊α2(2m+1)⌋-1, m∈N and α≥1,(60)-π≤π2αα2m+12+1-s-α2m+12+2α2+1+s≤-π2.This follows from (15). Thus ∀s=⌊α2(m+3/4)⌋+2,…,⌊α2(2m+1)⌋-1, α≥1, and for any odd number m we have(61)-1≤cos⁡πα2m+12+1-s2α-πα2m+12+2α2+1+s2α≤0.In view of (57), (59), and (61), for any odd number m, the following inequality is satisfied:(62)∑s=α2m+3/4+2α22m+1-1sin⁡πα2m+12+1-sα+sin⁡πα2m+12+2α2+1+sα≤0.Now we see that ∀m∈N(63)sin⁡πα2m2+1α+…+sin⁡πα2m+22α=∑s=1α2m+3/4-1sin⁡πα2m+12-sα+sin⁡πα2m+12+sα+∑s=α2m+3/4+2α22m+1-1sin⁡πα2m+12+1-sα+sin⁡πα2m+12+2α2+1+sα+sin⁡πα2m2+1α+sin⁡πα2m2+2α+sin⁡πα2m+12-α2m+3/4α+sin⁡πα2m+12α+sin⁡πα2m+12+α2m+3/4α+…+sin⁡πα2m+12+2α2+α2m+3/4+2α+sin⁡πα2m+22α,which follows from the relationship(64)α2m+12-α22m+1+2<α2m2+2m+1-2m-1+3=α2m2+3<α2m2+4,α2m+12-α22m+1+2>α2m2+1≥α2m2+1,andα2m+22-α2m+12+2α2-α22m+1-1+1≤1.Note that, for some numbers m or α, some components will cease to exist in formula (63). As an example, let α=m=1. Then there are not ∑s=1⌊α2(m+3/4)⌋-1‍{…},∑s=⌊α2(m+3/4)⌋+2⌊α2(2m+1)⌋-1‍{…}, sin⁡π⌊α2m2⌋+2/α, sin⁡π⌊α2(m+2)2⌋/α in formula (63). However, an estimation of the number of components of (63) shall be sufficient for further consideration. Denote by X the set(65)α2m2+1,α2m2+2,α2m+12-α2m+34,α2m+12,α2m+12+α2m+34,…,α2m+12+2α2+α2m+34+2,α2m+22.We calculate(66)∀m∈N2α2+1<α2m+12+2α2+α2m+34+2-α2m+12-α2m+34+1<2α2+5and thus(67)X<2α2+10.Note that ∀α≥1 and ∀m∈Oddnumbers(68)sin⁡πα2m2+1α+…+sin⁡πα2m+22α≤2α2+10≤12α2.This follows from (56), (62), (63), and (67).
Note that(69)S′≤12α2∑m=mn′αm∈Oddnumbersmp′α-2cα2m+12≤12α2cα2mn′α+12+cα2mn′α+32+cα2mn′α+52+…+cα2mp′α-12≤12α2∑t=1∞cαmn′α+2t-1α2≤12α2∑t=1∞cn+2t-1α2=12α2∑t=1∞cn+2t-1α2n+2t-1α2n+2t-1α2≤12α2supt≥1cn+2t-1α2n+2t-1α2∑t=1∞1n+2t-1α2.This follows from (27), (35), (50), and (68). After considering that
∀ t ≥ 1 ⌊ ( n + ( 2 t - 1 ) α ) 2 ⌋ ≥ n we obtain (70)supt≥1⁡cn+2t-1α2n+2t-1α2≤supk≥n⁡kck.Thus(71)S′≤12supk≥nkckα2∑t=1∞1n+2t-1α2-1≤12supk≥nkck∑t=1∞14t2-1≤2π2supk≥nkck.This follows from (23). In view of (45) and (71) the following inequality is satisfied:(72)S≤32supk≥nkck.On the other hand(73)S=∑k=nα2mn′′2αcksin⁡πkα+∑k=α2mp′′2αpcksin⁡πkα+S′′,whereS′′=∑m=mn′′αm∈Evennumbersmp′′α-2∑z=α2m2+1α2m+1/22+∑z=α2m+1/22+1α2m+12+∑z=α2m+12+1α2m+3/22+∑z=α2m+3/22+1α2m+22czsin⁡πzα.. We see that(74)S≥-∑k=nα2mn′′2αcksin⁡πkα-∑k=α2mp′′2αpcksin⁡πkα+S′′≥-12supk≥nkck+S′′.This follows from (30) and (43). Note that for any even number m and ∀α≥1(75)cα2m+12∑z=α2m2+1α2m+12sin⁡πzα≤∑z=α2m2+1α2m+12czsin⁡πzαwhich follows from the relationship(76)sin⁡πα2m2+1α≥0,…,sin⁡πα2m+12α≥0. On the other hand, ∀α≥1 and for any even number m and, we have(77)cα2m+12∑z=α2m+12+1α2m+22sin⁡πzα≤∑z=α2m+12+1α2m+22czsin⁡πzα,which follows from the relationship(78)sin⁡πα2m+12+1α≤0,…,sin⁡πα2m+22α≤0.Thus(79)S′′≥cα2m+12∑z=α2m2+1α2m+22sin⁡πzα.This follows from (75), (77). Note that for s=1,2,…,⌊α2(m+3/4)⌋-1 and for any even number m(80)0≤sin⁡πα2m+12-s+α2m+12+s2α≤1(81)0≤cos⁡πα2m+12-s-α2m+12+s2α≤1.This follows from (52) and (54). Hence if m is an even number, then(82)∑s=1α2m+3/4-1sin⁡πα2m+12-sα+sin⁡πα2m+12+sα≥0.This follows from (51), (80), and (81). Note that for s=⌊α2(m+3/4)⌋+2,…,⌊α2(2m+1)⌋-1 and for any even number m(83)sin⁡πα2m+12+1-s+α2m+12+2α2+1+s2αand(84)cos⁡πα2m+12+1-s-α2m+12+2α2+1+s2αbelong to [-1,0]. This follows from (58) and (60). In view of (57), (83), and (84) and for any even number m, the following inequality is satisfied: (85)∑s=α2m+3/4+2α22m+1-1sin⁡πα2m+12+1-sα+sin⁡πα2m+12+2α2+1+sα≥0.Hence, if m is an even number, then(86)∑α2m2+1α2m+22sin⁡πzα≥-2α2-10≥-12α2.This follows from (63), (67), (82), and (85). Thus(87)S′′≥-12α2∑m=mn′′αm∈Evennumbersmp′′α-2cα2m+12.This follows from (79) and (86). In analogy with (69) and (71) we have(88)12α2∑m=mn′′αm∈Evennumbersmp′′α-2cα2m+12≤2π2supk≥nkck.This follows from (23), (28), and (36). In view of (87) and (88), the following inequality is satisfied:(89)S′′≥-2π2supk≥nkck.Hence(90)S≥-32supk≥nkck.This follows from (74) and (89). Thus(91)Forpn≥7wehave:∑k=npcksin⁡πkα≤32supk≥nkck.This follows from (72) and (90). In view of (25) and (91) the following inequality is satisfied:(92)∀p≥n∀α∈1,n∑k=npcksin⁡πkα≤49supk≥nkck.
Case 3:(93)n≤α≤p.In view of (24) and (25) there is one case to consider, namely, p/n≥7. Then(94)∃r∈Nthatr>nandn≤r-1≤α2≤r≤p.Note that(95)∑k=npcksin⁡πkα=∑k=nr-1cksin⁡πkα+∑k=rpcksin⁡πkα.We now give an estimate for ∑k=nr-1‍cksin⁡πk/α. Note that 0<πk/α<π and 0≤sin⁡πk/α≤πk/α for k=n,…,r-1. This follows from (94). Hence(96)0≤∑k=nr-1cksin⁡πkα≤∑k=nr-1ckπkα≤πsupk≥nkck∑k=nr-11kα≤πsupk≥nkck∑k=nr-11kr-1≤2πsupk≥nkck.This follows from (9) and (94).
We now give an estimate for∑k=rp‍cksin⁡πk/α. There are two cases to consider:
Case 3’:p/r<7.
Then, by (94) and (96), we have(97)∑k=npcksin⁡πkα≤∑k=nr-1cksin⁡πkα+∑k=rpcksin⁡πkα≤2π+49supk≥nkck.
Case 3”:p/r≥7.
Note that the replacement ofn with r in (23) and (26) gives us(98)∑k=rpcksin⁡πkα≤32supk≥nkckand thus in case 3 we proved that ∀n,p, where n≤p, and ∀α∈[n,p](99)∑k=npcksin⁡πkα≤56supk≥nkck. This follows from (96), (97), and (98). In view of cases 1, 2, and 3 we obtain(100)∀α≥1∀n,pwheren≤p:∑k=npcksin⁡πkα≤max⁡2π,49,56supk≥nkck.This follows from (11), (92), and (99). If limn→∞ncn=0, then
∀ ϵ > 0   ∃ M   ∀ n > M   s u p k ≥ n { k c k } < ϵ, so(101)∀n,pwheren≤p:∑k=npcksin⁡πkα≤56ϵforallα≥1provided that n>M. This finishes the proof of the sufficient condition. This completes the proof of Theorem 6.Remark 7.
If{ck}k=1∞⊂R+ is nonincreasing, then ∑k=1∞‍cksin⁡k2θ converges uniformly in θ∈[0,π] if and only if ∑k=1∞‍ck is convergent.Proof.
Proof of the sufficient condition is obvious.
Proof of the necessary condition. Let(102)∑k=1∞cksin⁡πk28be a convergent series. Note that for any even number m we have(103)∑k=4m4m+2-1cksin⁡πk28=c4m+1sin⁡π2m2+m+18+c4m+2sin⁡π2m2+2m+48+c4m+3sin⁡π2m2+3m+98+c4m+5sin⁡π2m2+5m+258+c4m+6sin⁡π2m2+6m+368+c4m+7sin⁡π2m2+7m+498≥c4m+3sin⁡π8+sin⁡9π8+c4m+5sin⁡4π8+sin⁡25π8>12c4m+5.Thus(104)∑k=8∞cksin⁡k2π8=∑m∈Evennumbersm≥2∑k=4m4m+7cksin⁡k2π8>∑m∈Evennumbersm≥2c4m+52.This follows from (103). After considering that ∑k=8∞‍cksin⁡k2π/8 is convergent, we obtain the convergent series c13+c21+c29+…+c8r+5+…. This follows from (104). After considering that {ck}k=1∞ is nonincreasing, we have that
∀ s = 0,1 , … , 7 ∑ r = 1 ∞ ‍ c 8 r + 5 + s is convergent and ∑r=1∞‍cr is convergent.



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*Source: 1342189-2019-03-20.xml*