Question 9.10: Discuss the convergence properties of the two series (i) ∑si...

Discuss the convergence properties of the two series

(i)sinnxn2, (ii)sinnxn.(i) \sum{\frac{\sin nx}{n^{2}}} ,  (ii) \sum{\frac{\sin nx}{n}}.

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(i) Since |sin nx| ≤ 1 for all xRx ∈ \mathbb{R} and 1n2\sum{\frac{1} {n^{2}}} is convergent, the series sinnxn2\sum{\frac{\sin nx}{n^{2}}} is uniformly convergent on R\mathbb{R} by the M test.

(ii) If  sinnxnMn\left|\frac{\sin nx}{n}\right| ≤ M_{n} for all x in some interval I, then Mncn,M_{n} ≥ \frac{c}{n}, where c=supxIsinnx>0.c = \underset{x∈I}{\sup} |\sin nx| > 0. Since 1n\sum{\frac{1} {n}} is divergent, we see that the M-test is not applicable on any interval in R.\mathbb{R}. We shall use the Dirichlet test to prove uniform convergence on any interval [a,b](2mπ,(2m+1)π), mZ.[a, b] ⊆ (2mπ, (2m + 1)π),  m ∈ \mathbb{Z}. The proof relies on the equality

k=1nsinkx=cos12xcos(n+12)x2sin12x, x2mπ,mZ,\sum\limits_{k=1}^{n}{\sin kx} = \frac{\cos \frac{1}{2} x − \cos(n + \frac{1}{2})x}{2 \sin \frac{1}{2} x} ,  x ≠ 2mπ, m ∈ Z,        (9.16)

which can be proved by induction on n. Setting uk(x)=sinkxu_{k}(x) = \sin kx and Un(x)=Σk=1n uk(x),U_{n}(x) = \Sigma_{k=1} ^{n}  u_{k}(x), we see that

Un(x)cos12x+cos(n+12)x2sin12x|U_{n}(x)| ≤ \frac{\left|\cos \frac{1}{2}x\right| + \left|\cos(n + \frac{1}{2})x\right|}{2 \left|\sin \frac{1}{2} x\right|}

1sin12x≤ \frac{1}{\left|\sin \frac{1}{2} x\right|}

max{1sin12a,1sin12b}≤ \max \left\{\frac{1}{\left|\sin \frac{1}{2} a\right|}, \frac{1}{\left|\sin \frac{1}{2}b\right|}\right\}  for all nN, x[a,b].n ∈ \mathbb{N},  x ∈ [a, b].

With vn(x)=1/nv_{n}(x) = 1/n on [a, b], we obtain a decreasing sequence which converges (uniformly) to 0. By Dirichlet’s test the series sinnxn\sum{\frac{\sin nx}{n}} converges uniformly on [a, b].

Using Fourier expansions (see [CAR], for example), it can be shown that sinnxn\sum{\frac{\sin nx}{n}} converges pointwise on R\mathbb{R} to a function which is periodic in 2π, and which is defined on [−π, π] by

S(x)={π+x2,x[π,0)0,x=0πx2,x(0,π].S(x)= \begin{cases} − \frac{π + x}{2}, & x ∈ [−π, 0) \\ 0, & x= 0 \\ \frac{π − x}{2}, & x ∈ (0, π]. \end{cases}

Note that x=2mπ, mZ,x = 2mπ,  m ∈ \mathbb{Z}, are points of (jump) discontinuity for the function S. Hence the series sinnxn\sum{\frac{\sin nx}{n}} of continuous terms cannot converge uniformly in any interval containing such points.

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