Friday, 8 April 2016

real analysis - Prove limlimitsntoinftyfrac1sqrtnleft|sumlimitsnk=1(1)ksqrtkright|=frac12



I'm trying to show that limnxn=limn1n|nk=1(1)kk|=12.


Assuming limnxn=x exists, we have



x2n=2n1(x2n1)+2n2n



Letting n,




x=x+1


x=12



But I'm stuck on proving the existence of limxn. Any idea?






Update: I just solved the problem using sandwich theorem + integral test. Still, I would like to see a continuation of my initial idea, i.e. proving





{x2n} is monotonically increasing (similarly, {x2n+1} is monotonically decreasing)



Answer



limn1n|nk=1(1)kk|=limn12nnk=112k1+2k



122nnk=112k<12nnk=112k1+2k<122nnk=112k1<122n(1+n1k=112k)



With Riemann we get:




12nnk=112k=12nnk=11k/n1210dxx=x|10=1



And therefore the confirmation of the claim.









About the monotonie.




sn:=nk=112k1+2k



Assumption about the monotonie: sn2n is strictly increasing.



sn2n<sn+12n+2 leads to



(12n12n+2)sn<12n+212n+1+2n+2 and therefore to



sn<an:=2n22n+2+2n2n+2+2n+1




For n=1 it’s o.k. . Assume it’s correct for n . ()



Then we have to show that it's also correct for nn+1 .



sn+1<2n22n+2+2n2n+2+2n+1+12n+1+2n+2<an+1



The first inequation follows from the assumption () and the second inequation can be proved by some transformations, 2n replaced by x and x0:



(x(x+2+x)+2)(x+4+x+3)<




<x+2(x+4+x+2)(x+2+x+1)



Transformations:



(x(x+2+x)+2)(x+4+x+2)+(x(x+2+x)+2)(x+3x+2)<
<x+2(x+4+x+2)(x+2+x+1)



(x(x+2+x)+2)(x+3x+2)<(x+2(x+2+x+1)(x(x+2+x)+2))(x+4+x+2)



(xx+2+x+2)(x+3x+2)<(x+2x+4+x+2)(x+1x)




This is true because of:



xx+2+x+2<x+2x+4+x+2



x+3x+2<x+1x



Note: x+1+ax+a  is decreasing by growing a>x


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