Respuesta :

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Given that [tex]f^{(n)}(0)=(n+1)![/tex], we have for [tex]f(x)[/tex] the Taylor series expansion about 0 as

[tex]f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n[/tex]

Replace [tex]n+1[/tex] with [tex]n[/tex], so that the series is equivalent to

[tex]f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}[/tex]

and notice that

[tex]\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}[/tex]

Recall that for [tex]|x|<1[/tex], we have

[tex]\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}[/tex]

which means

[tex]f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}[/tex]
[tex]\implies f(x)=\dfrac1{(1-x)^2}[/tex]

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