Question 1091635
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We will use the fact that

{{{(1+y)^(-n)}}}{{{""=""}}}{{{sum((-1)^k*(matrix(2,1,n+k-1,k))y^k,k=0,infinity)}}} converges for {{{abs(y)<1}}},

and where {{{(matrix(2,1,p,q))}}} is the number of combinations
of p things taken q at a time or {{{p!/(q!(p-q)!)}}}

{{{(2+x)^(-4)}}}{{{""=""}}}{{{(x+2)^(-4)}}}{{{""=""}}}{{{(x(1+2/x)^"")^(-4)}}}{{{""=""}}}{{{x^(-4)(1+2/x)^(-4)}}}{{{""=""}}}

{{{x^(-4)sum((-1)^k*(matrix(2,1,n+k-1,k))(2/x)^k,k=0,infinity)}}} converges for {{{abs(2/x)<1}}}

Simplifying,

{{{x^(-4)sum((-1)^k*(matrix(2,1,n+k-1,k))(2^k/x^k),k=0,infinity)}}} converges for {{{abs(x/2)>1}}}

{{{x^(-4)sum((-1)^k*(matrix(2,1,n+k-1,k))(2^k*x^(-k)),k=0,infinity)}}} converges for {{{abs(x)>2}}}

{{{sum((-1)^k*(matrix(2,1,n+k-1,k))(2^k*x^(-k-4)),k=0,infinity)}}} converges for {{{abs(x)>2}}}

{{{sum((-1)^k*(matrix(2,1,n+k-1,k))(2^k/x^(k+4)),k=0,infinity)}}} converges for {{{abs(x)>2}}}

Edwin</pre></b></font>