Saturday, 2 January 2016

Convergence of Series suminftyn=1left[fracsinleft(fracn2+1nxright)sqrtnleft(1+frac1nright)nright]




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 [2kπε,2(k+1)πε].



Can I just say sin(n2+1nx)n1n0 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]=en=1[sin(nx)n+xcos(nx)nn+o(1nn)]



which converges since



n=1sin(nx)n converges by Dirichlet's test




n=1xcos(nx)nn converges absolutely by comparison with n=11nn


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