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lim-n-r-1-n-1-2-r-where-greatest-integer-finction-




Question Number 213511 by universe last updated on 07/Nov/24
             lim_(n→∞)    [Σ_(r=1) ^n (1/2^r )]     where [•] greatest integer finction
limn[nr=112r]where[]greatestintegerfinction
Answered by issac last updated on 07/Nov/24
Σ_(h=1) ^M  2^(−h) ≤1  lim_(M→∞) Σ_(h=1) ^M  2^(−h) ≤1  but Σ_(h=1) ^∞  2^(−h) =1  cus 0.99999......=1  ⌊lim_(M→∞)  Σ_(h=1) ^M  2^(−h) ⌋=1
Mh=12h1limMMh=12h1buth=12h=1cus0.99999=1limMMh=12h=1
Commented by mr W last updated on 07/Nov/24
the question is  lim_(n→∞) ⌊×⌋=?  not ⌊lim_(n→∞) ×⌋=?
thequestionislimn×=?notlimn×=?
Answered by lepuissantcedricjunior last updated on 09/Nov/24
lim_(n→+∞) [Σ_(r=1) ^n (1/2^r )]=lim_(n→+∞) [−1+Σ_(r=0) ^n (1/2^r )]  =lim_(n→+∞) [−1+((1−((1/2^(n+1) )))/(1−(1/2)))]  =lim_(n→+∞) [−1+2(1−2^(−(n+1)) )]  =[−1+2]=1
limn+[nr=112r]=limn+[1+nr=012r]=limn+[1+1(12n+1)112]=limn+[1+2(12(n+1))]=[1+2]=1
Commented by mr W last updated on 09/Nov/24
wrong!
wrong!
Answered by mr W last updated on 07/Nov/24
Σ_(r=1) ^n (1/2^r )=(((1/2)(1−(1/2^n )))/(1−(1/2)))=1−(1/2^n )<1, but >0  ⇒[Σ_(r=1) ^n (1/2^r )]=0  ⇒lim_(n→∞) [Σ_(r=1) ^n (1/2^r )]=lim_(n→∞) 0=0 ✓
nr=112r=12(112n)112=112n<1,but>0[nr=112r]=0limn[nr=112r]=lim0n=0
Commented by issac last updated on 07/Nov/24
Damn i was wrong
Damniwaswrong
Commented by universe last updated on 07/Nov/24
thanks sir
thankssir
Answered by Berbere last updated on 07/Nov/24
[x]<x  ⇒0≤[Σ_(r=1) ^n (1/2^r )]≤Σ_(r=1) ^n (1/2^r )=1−((1/2))^n <1  ⇒∀n∈N  [Σ(1/2^r )]=0
[x]<x0[nr=112r]nr=112r=1(12)n<1nN[Σ12r]=0
Commented by universe last updated on 07/Nov/24
thank you so much sir
thankyousomuchsirthankyousomuchsir

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