Consider the series ∞∑n=1[sin(n2+1nx)√n(1+1n)n] . Find all points at which the series
is convergent. Find all intervals that the series is uniformly
convergent.
I know that I need to use Dirichlet and/or Abel Criterion to show this.
My first attempt was to consider the argument of the sum as product of three sequences of functions, fn,gn,hn and perform Dirichlet/Abel twice. Towards that end I tried to break up
sin(n2+1nx) using sin(α+β)=sinαcosβ+sinβcosα to produce
sin(nx+xn)=sin(nx)cos(xn)+sin(xn)cos(nx) and use the fact that sin(nx)≤1|sin(x2)|, but then I don't know what to do with the sin(xn), cos(xn), and cos(nx). Previously, when using this method we showed uniform convergence on compact intervals like [2kπ−ε,2(k+1)π−ε].
Can I just say sin(n2+1nx)√n≤1√n→0 as n→∞ ?
Also, I know that limn→∞(1+1n)n=e.
i.e. Help :) Thank you in advance
Answer
Note that
sin(n2+1nx)=sin(nx+1nx)=sin(nx)cos(1nx)+cos(nx)sin(1nx)=
=sin(nx)+xcos(nx)n+o(1n)
(1+1n)n=enlog(1+1n)=e1+1n+o(1n)=e(1+1n+o(1n))
thus
∞∑n=1[sin(n2+1nx)√n(1+1n)n]=e∞∑n=1[sin(nx)√n+xcos(nx)n√n+o(1n√n)]
which converges since
∑∞n=1sin(nx)√n converges by Dirichlet's test
∑∞n=1xcos(nx)n√n converges absolutely by comparison with ∑∞n=11n√n
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