Let {an} be the decreasing sequence of positive terms. Suppose ∑n=1∞ansin(nx) converges uniformly and let the first n partial sum of the given series be Sn(x)=∑r=1narsin(rx) , By the hypothesis Sn(x) converges uniformly.Now by Cauchy's criteria of uniform convergence, for every ϵ>0,∃N∈N such that n≥N,m≥N for sufficiently large m,n⟹
∣Sn(x)−Sm(x)∣<ϵ,∀x∈R ⟺ ∣∑r=m+1narsin(rx)∣<∑r=m+1n∣arsin(rx)∣ (since, from general triangle inequality),thus
∣∑r=m+1narsin(rx)∣<∑r=m+1n∣arsin(rx)∣=∑r=m+1n∣ar∣∣sin(rx)∣ .We also know that, always ∣sin(rx)∣<1 ⟺∑r=m+1n∣ar∣∣sin(rx)∣<∑r=m+1n∣ar∣=∑r=m+1nar (since, ar>0,∀r∈N .Hence choose ϵ=∑r=m+1nar⟺ϵ=∑r=m+1nar>an(n−m) (∵{an} is decreasing sequence),thus an→0, as n→∞ (since, ϵ is finite positive real number).
Hence, we are done.
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