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Prove-that-for-all-complex-such-as-z-lt-1-n-1-z-n-z-n-1-2-n-1-nz-n-z-n-1-0-




Question Number 89937 by ~blr237~ last updated on 20/Apr/20
Prove that for all complex such as ∣z∣<1=  Σ_(n=1) ^∞ (z^n /((z^n −1)^2 )) +Σ_(n=1) ^∞  ((nz^n )/(z^n −1)) = 0
Provethatforallcomplexsuchasz∣<1=n=1zn(zn1)2+n=1nznzn1=0
Commented by mathmax by abdo last updated on 20/Apr/20
we have for ∣u∣<1   Σ_(p=0) ^∞ u^p   =(1/(1−u)) ⇒Σ_(p=1) ^∞ pu^(p−1)  =(1/((1−u)^2 ))  u=z^n  ⇒(1/((1−z^n )^2 )) =Σ_(p=1) ^∞ p(z^n )^(p−1)  =Σ_(p=1) ^∞  p (z^(p−1) )^n  ⇒  Σ_(n=1) ^∞  (z^n /((z^n −1)^2 )) =Σ_(n=1) ^∞ z^n Σ_(p=1) ^∞ p (z^(p−1) )^n   =Σ_(p=1) ^∞  pΣ_(n=1) ^∞ (z.z^(p−1) )^n  =Σ_(p=1) ^∞ pΣ_(n=1) ^∞  z^(pn)   from another side  Σ_(n=1) ^∞  ((nz^n )/(z^n −1)) =−Σ_(n=1) ^∞  nz^n  Σ_(p=0) ^∞  z^(np)  =−Σ_(p=1) ^∞ p z^p  Σ_(n=0) ^∞  z^(pn)   =−Σ_(p=1) ^∞  p Σ_(n=0) ^∞ z^(p(n+1) ) =−Σ_(p=1) ^∞  p Σ_(n=1) ^∞  z^(pn)  ⇒  Σ_(n=1) ^∞  (z^n /((z^n −1)^2 )) +Σ_(n=1) ^∞  ((nz^n )/(z^n −1)) =0
wehaveforu∣<1p=0up=11up=1pup1=1(1u)2u=zn1(1zn)2=p=1p(zn)p1=p=1p(zp1)nn=1zn(zn1)2=n=1znp=1p(zp1)n=p=1pn=1(z.zp1)n=p=1pn=1zpnfromanothersiden=1nznzn1=n=1nznp=0znp=p=1pzpn=0zpn=p=1pn=0zp(n+1)=p=1pn=1zpnn=1zn(zn1)2+n=1nznzn1=0

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