Monday, 19 November 2018

sequences and series - Prove that xmapstosuminftyn=0fn(x)=suminftyn=0fracx2(1+x2)n does not converge uniformly on [1,1]



For each n, we define fn:RR
xfn(x)=x2(1+x2)n
We consider the function
f:RR

xf(x)=n=0fn(x)=n=0x2(1+x2)n



I want to show that the series does not converge uniformly on [1,1] but I'm finding it difficult to do that.



First of all, I considered the Weierstrass M test.



MY TRIAL



|fn(x)|=|x2(1+x2)n|1(1+x2)n,x[1,1],nN




I'm also thinking that the βn=sup approach could be very helpful too!


Answer



The convergence is not uniform.



For x \ne 0 this is a geometric series:



\sum^{\infty}_{n=0}\frac{x^2}{(1+x^2)^n} = x^2\cdot \frac{1}{1-\frac1{1+x^2}}= x^2+1



and for x = 0 the sum is 0.




Therefore



f(x) = \begin{cases} x^2+1, &\text{ if } x \ne 0\\ 0, &\text{ if }x = 0 \end{cases}



Since f is not continuous, the convergence cannot be uniform.


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