Question Number 100968 by bachamohamed last updated on 29/Jun/20

      Σ_(k=1) ^∞  (x+k)^(1/2^(k+1) ) =?      x>0

Answered by mathmax by abdo last updated on 29/Jun/20

S =Σ_(n=1) ^∞ (x+n)^(1/(2^(n+1)  ))  =Σ_(n=1) ^∞  U_n    we have U_n >0 ⇒  (U_(n+1) /U_n ) =(((x+n+1)^(1/2^(n+2) ) )/((x+n)^(1/2^(n+1) ) )) =(({(x+n+1)^(1/2^(n+1) ) }^(1/2) )/({(x+n)^(1/2^n ) }^(1/2) )) =(√(((x+n+1)^(1/2^(n+1) ) )/((x+n)^(1/2^n ) )))  =(√((√(x+n+1))×(((x+n+1)/(x+n)))^(1/2^n ) ))=(x+n+1)^(1/4) ×(((x+n+1)/(x+n)))^(1/2^(n+1) )   =(x+n+1)^(1/4) ×(1+(1/(x+n)))^(1/(2^(n+1)  )) ∼(x+n+1)^(1/4)  ×(1+(1/(2^(n+1) (x+n))))  =(x+n+1)^(1/4) +(((x+n+1)^(1/4) )/(2^(n+1) (x+n))) →+∞ ⇒this serie is divergent...!

Commented bybachamohamed last updated on 29/Jun/20

thank′s sur but   Σ_(k=1) ^n (x+k)^(1/2^(k+1) ) ={(x+1)^(1/2^2 ) +(x+2)^(1/2^3 ) +(x+3)^(1/2^4 ) ......(x+n)^(1/2^(n+1) )    ⇒ Σ_(k=1) ^(k=n) (x+k)^(1/2^(k+1) ) =(√(√((x+1)+(√((x+2)+(x+3)...+.....(√((x+n))))))))=1  ⇒ Σ_(k=1) ^∞ (x+k)^(1/2^(k+1) ) =(√(√((x+1)+(√((x+2)+(√((x+3)+(√((x+4)+........∞)))))))))=1   ⇒ serie is converge   pourqoi?

Commented bymaths mind last updated on 29/Jun/20

1st lign ⇏2nd lign  (√x)+(√y)≠(√(x+(√y)))

Commented bybachamohamed last updated on 29/Jun/20

no it′ s right just look

Commented bymathmax by abdo last updated on 29/Jun/20

sir you must prove this by recurrence your snswer is not clear...

Answered by mathmax by abdo last updated on 29/Jun/20

if you have another method post it sir bacha

Commented bybachamohamed last updated on 29/Jun/20

i am looking for other ways becaus    i have reached many contradiction  so i shared the idea with you

Commented bymathmax by abdo last updated on 30/Jun/20

nevermind sir you are always welcome...