f=z+z(z2โzโ2)3โ=z+z(z+1)(zโ2)3โ=z+z1โ(zโ21โโz+11โ)โนz+z(zโ2)1โโz(z+1)1โ=z+2(zโ2)1โโ2z1โโz1โ+z+11โ=zโ23โzโ1+4(2zโโ1)1โ+z+11โ
We remind the formula: a+ar+...+arn=a(1โr1โrn+1โ), โฃrโฃ<1.
We take the limit and get: a+ar+...+arn+...=1โraโ.
a). For โฃzโฃ<1 we use the formula and get:
โนzโ23โzโ1+4(2zโโ1)1โ+z+11โ=zโ23โzโ1โ41โ(1+2zโ+22z2โ+23z3โ...)+(1โz+z2โz3+z4+...)
b). We point out that 21โ<โฃzโฃ1โ<1. We rewrite the equality as:
โนzโ23โzโ1โ4(1โ2zโ)1โ+z(1+z1โ)1โ=zโ23โzโ1โ41โ(1โ2zโ+4z2โโ...)+z1โ(1โz1โ+z21โ+...)
c). We have: 21โ>โฃzโฃ1โ. Thus, we get:
โนzโ23โzโ1+2(1โz2โ)1โ+z(1+z1โ)1โ=zโ23โzโ1+21โ(1โz2โ+z24โ+...)+z1โ(1โz1โ+z21โ+...)
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