Respuesta :
[tex]f(x)=\displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n[/tex]
Note that, by the ratio test, this converges for
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)x^{n+1}}{(n+1)x^n}\right|=|x|<1[/tex]
Integrating once, you have
[tex]\displaystyle\int f(x)\,\mathrm dx=\int\sum_{n=0}^\infty(n+1)x^n\,\mathrm dx=\sum_{n=0}^\infty(n+1)\int x^n\,\mathrm dx[/tex]
[tex]\displaystyle\int f(x)\,\mathrm dx=\sum_{n=0}^\infty\frac{(n+1)x^{n+1}}{n+1}+C=\sum_{n=0}^\infty x^{n+1}+C[/tex]
Given that [tex]f(0)=0[/tex], it's clear that [tex]C=0[/tex], and so
[tex]\displaystyle\int f(x)\,\mathrm dx=\displaystyle\sum_{n=0}^\infty x^{n+1}=x\sum_{n=0}^\infty x^n=\frac x{1-x}[/tex]
Differentiating both sides recovers [tex]f(x)[/tex]:
[tex]f(x)=\dfrac{(1-x)(1)-x(-1)}{(1-x)^2}=\dfrac1{(1-x)^2}[/tex]
Note that, by the ratio test, this converges for
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)x^{n+1}}{(n+1)x^n}\right|=|x|<1[/tex]
Integrating once, you have
[tex]\displaystyle\int f(x)\,\mathrm dx=\int\sum_{n=0}^\infty(n+1)x^n\,\mathrm dx=\sum_{n=0}^\infty(n+1)\int x^n\,\mathrm dx[/tex]
[tex]\displaystyle\int f(x)\,\mathrm dx=\sum_{n=0}^\infty\frac{(n+1)x^{n+1}}{n+1}+C=\sum_{n=0}^\infty x^{n+1}+C[/tex]
Given that [tex]f(0)=0[/tex], it's clear that [tex]C=0[/tex], and so
[tex]\displaystyle\int f(x)\,\mathrm dx=\displaystyle\sum_{n=0}^\infty x^{n+1}=x\sum_{n=0}^\infty x^n=\frac x{1-x}[/tex]
Differentiating both sides recovers [tex]f(x)[/tex]:
[tex]f(x)=\dfrac{(1-x)(1)-x(-1)}{(1-x)^2}=\dfrac1{(1-x)^2}[/tex]
The Maclaurin series for f is found by using the ratio test, integration, and differentiation is given below.
[tex]f(x) = \dfrac{1}{(1-x)^2}[/tex]
What is a series?
A series is a sum of sequence terms. That is, it is a list of numbers with adding operations between them.
The Maclaurin series for f will be given below.
[tex]f(x) = \sum_{n=0}^{\infty} \dfrac{f^{(n)} (0)}{n!} \ x^n = \sum_{n=0}^{\infty} \dfrac{(n+1)!}{n!} \ x^n = \sum_{n=0}^{\infty} (n+1) \ x^n[/tex]
By the ratio test, this converges for
[tex]\displaystyle \lim_{n \to \infty} \left| \dfrac{(n+2)x^{n+1}}{(n+1)x^n} \right| = |x| < 1[/tex]
Integrating once, then we have
[tex]\rm \int f(x)dx = \int \sum_{n=0}^{\infty} (n+1)x^n dx = \sum_{n=0}^{\infty} (n+1 ) \int x^ndx\\\\\\\int f(x)dx = \int \sum_{n=0}^{\infty} \dfrac{(n+1 ) x^{n+1}}{n+1} +C = \sum_{n=0}^{\infty} x^{n+1} +C[/tex]
Given that f(0) = 0, then the value of C will be 0. Then we have
[tex]\rm \int f(x)dx = \sum_{n=0}^{\infty} x^{n+1} = x\sum_{n=0}^{\infty} x^n = \dfrac{x}{1-x}[/tex]
On differentiating, we have
[tex]f(x) = \dfrac{(1-x)*1-x(-1)}{(1-x)^2} = \dfrac{1}{(1-x)^2}[/tex]
More about the series link is given below.
https://brainly.com/question/10813422