Answer :
Answer:
A.
[tex]\mathbf{f(x)=C + \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ x^{14n +8}}{(2n+1)(14n+8)}}[/tex]
For convergence since |x| > 1
The radius of convergence R = 1
B.
[tex]\mathbf{f(x) = C + \sum \limits ^{\infty}_{n =0} \dfrac{(-1)^n \ x^{n+9}} {(n+1) (x^{n+9})}}[/tex]
For convergence since |x| < 1
The radius of convergence R = 1
Step-by-step explanation:
A.
Given that:
[tex]f(x) = \int tan^{-1} (x^7) \ dx[/tex]
Let recall that for Power series of tan⁻¹ (x)
[tex]tan^{-1} (x) = \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n x^{2n+1}}{(2n+1)}[/tex]
Then [tex]tan^{-1} (x^7) = \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n (x^7)^{2n+1}}{(2n+1)}[/tex]
[tex]tan^{-1} (x^7) = \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ x^{14n+7}}{(2n+1)}[/tex]
Thus;
[tex]f(x) =\int tan^{-1} (x^7) \ dx = \int \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ x^{14n+7}}{(2n+1)}[/tex]
[tex]\implies \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n }{(2n+1)} \int x^{14n+7} \ dx[/tex]
[tex]\mathbf{f(x)=C + \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ x^{14n +8}}{(2n+1)(14n+8)}}[/tex]
For convergence since |x| > 1
The radius of convergence R = 1
B.
[tex]\int x^7 \ In (1 + x) \ dx[/tex]
Recall that for power series of,
[tex]In(1+x) = \sum \limits ^{\infty}_{n = 0} \dfrac{(-1)^n \ x^{n+1}}{n +1}[/tex]
Thus;
[tex]x^7 \ In (1+x) = x^7 \sum \limits ^{\infty}_{n =0} \dfrac{(-1)^n \ x^{n+1} }{n+1}[/tex]
[tex]\implies \sum \limits ^{\infty}_{n =0} \dfrac{(-1)^n \ x^{n+8} }{n+1}[/tex]
[tex]f(x) = \int x^7 \ In (1+x) \ dx = \int \sum \limits ^{\infty}_{n =0} \dfrac{(-1)^n \ x^{n+8} }{n+1} \ dx[/tex]
[tex]=\sum \limits ^{\infty}_{n =0} \dfrac{(-1)^n}{n+1} \int \ x^{n+8} \ dx[/tex]
[tex]\mathbf{f(x) = C + \sum \limits ^{\infty}_{n =0} \dfrac{(-1)^n \ x^{n+9}} {(n+1) (x^{n+9})}}[/tex]
For convergence since |x| < 1
The radius of convergence R = 1