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 lim.




i.e. Help :) Thank you in advance


Answer



Note that



\sin \left( \frac{n^2+1}{n}x\right)=\sin \left( nx+\frac{1}{n}x\right)=\sin(nx)\cos\left(\frac{1}{n}x\right)+\cos(nx)\sin\left(\frac{1}{n}x\right)=



=\sin(nx)+\frac{x\cos(nx)}{n}+o\left(\frac1n\right)



\left( 1+\frac{1}{n}\right)^n=e^{n\log\left( 1+\frac{1}{n}\right)}=e^{1+\frac1n+o\left(\frac1n\right)}=e\left({1+\frac1n+o\left(\frac1n\right)}\right)




thus



\displaystyle{ \sum_{n=1}^{\infty} \left[ \frac{\sin \left( \frac{n^2+1}{n}x\right)}{\sqrt{n}}\left( 1+\frac{1}{n}\right)^n\right]}= e\sum_{n=1}^{\infty}\left[\frac{\sin(nx)}{\sqrt n}+\frac{x\cos(nx)}{n\sqrt n}+o\left(\frac1{n\sqrt n}\right)\right]



which converges since



\sum_{n=1}^{\infty}\frac{\sin(nx)}{\sqrt n} converges by Dirichlet's test




\sum_{n=1}^{\infty}\frac{x\cos(nx)}{n\sqrt n} converges absolutely by comparison with \sum_{n=1}^{\infty}\frac{1}{n\sqrt n}


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