Question #f2365

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Answer 1 · 2018-02-16T00:39:16

#1/(1-x)^n = sum_(k=n-1)^oo ((k),(n-1))x^(k-n+1) #

converging at least for #x in (-1,1)#

#sum_(k=0)^oo x^k = 1/(1-x)#

converging for #x in (-1,1)#.

Inside the interval of convergence we can differentiate term by term and obtain a series with the same radius of convergence, so:

#d/dx 1/(1-x) = sum_(k=0)^oo d/dx x^k #

#1/(1-x)^2 = sum_(k=1)^oo kx^(k-1) #

iterating the process we have:

#2/(1-x)^3 = sum_(k=2)^oo k(k-1)x^(k-2) #

#(3!)/(1-x)^4 = sum_(k=3)^oo k(k-1)(k-2)x^(k-3) #

and in general:

#((n-1)!)/(1-x)^n = sum_(k=n-1)^oo (k!)/((k-n+1)!)x^(k-n+1) #

#1/(1-x)^n = 1/((n-1)!)sum_(k=n-1)^oo (k!)/((k-n+1)!)x^(k-n+1) #

or using the binomial coefficient:

#((k),(n-1)) = (k!)/((k-n+1)!(n-1)!)#

#1/(1-x)^n = sum_(k=n-1)^oo ((k),(n-1))x^(k-n+1) #

always converging at least for #x in (-1,1)#