XPost: sci.math   
   From: PointedEars@web.de   
      
   Ross Finlayson wrote:   
   > Then, I would argue that it demands of act of deduction after induction,   
   > the inference, to truly establish that the limit exists.   
      
   Not really.  The proof is simple, straightforward and does not require   
   induction:   
      
              s = ∑_{k=0}^∞ x^k = x^0 + x^1 + x^2 + ... = 1 + x + x^2 + ...   
   ⇔     s - 1 = x + x^2 + x^3 + ... = s x   
   ⇔   s - s x = 1   
   ⇔ (1 - x) s = 1   
   ⇔         s = 1/(1 - x), |x| < 1.   
      
   We have to require |x| < 1 because if x = -1 then the partial sums alternate   
   between 1 and -1, so the series does not converge; and if x >= 1 it does not   
   converge either as it approaches positive infinity (which is also indicated   
   as then in the derived formula for x = 1 we would divide by zero).  And if   
   x < -1, it certainly does not converge as the partial sums are alternating,   
   with the modulus of each partial sum greater than that of the previous one   
   (|x^{n+1}| > |x^n| ⇔ |x| > 1).   
      
   You can also use the Cauchy--Hadamard theorem according to which the radius   
   of convergence of a power series   
      
     s = ∑_{n=0}^∞ a_n (z - c)^n; a_n, z, c ∈ ℂ   
      
   is given by   
      
     r = 1/lim sup_{n → ∞}(|a_n|^{1/n}),   
      
   so that the bounds of the interval of convergence are c - r and c + r.   
      
   Since here a_n = 1 for all n, and c = 0, we find r = 1; but the interval of   
   convergence is open, z = x ∈ (-1, 1), because of the argument above.   
      
      
      
   > So, about relativity, where we may say that the only aspect of   
   > relativity theory as relative that _motion_ is relative,   
      
   Word salad.   
      
   > not including space itself or time   
      
   Nonsense.   
      
   > or for that matter the higher orders of acceleration, where for   
   > example Einstein gives a "the space" (absolute)   
      
   No.   
      
   > and separately the usual "spatial" and a "spacial" for Special   
   > a.k.a. Restricted relativity,   
      
   Einstein does not use "spacial".  Probably because that is not a word, but a   
   misspelling of "spatial", and Einstein wrote in German.   
      
   > which intends to include the L-principle   
   You are making up words but you do not understand the basics.   
      
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   PointedEars   
      
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