I'm guessing you're given
[tex]f(x)=\displaystyle\sum_{n=1}^\infty((-1)^n+1)(x-4)^n[/tex]
Note that for odd [tex]n[/tex], the corresponding term in the series is 0, so only the even terms matter. Let [tex]n=2k[/tex], for which [tex](-1)^{2k}+1=1+1=2[/tex]. Then
[tex]f(x)=\displaystyle2\sum_{k=1}^\infty(x-4)^{2k}[/tex]
Differentiating/integrating the power series gives
[tex]f'(x)=\displaystyle4\sum_{k=1}^\infty k(x-4)^{2k-1}[/tex]
[tex]f''(x)=\displaystyle4\sum_{k=1}^\infty k(2k-1)(x-4)^{2k-2}[/tex]
[tex]\displaystyle\int f(x)\,\mathrm dx=2\sum_{k=1}^\infty\frac{(x-4)}^{2k+1}}{2k+1}+C[/tex]
By the ratio test, ...
a. ... [tex]f(x)[/tex] converges for
[tex]\displaystyle\lim_{k\to\infty}\left|\frac{2(x-4)^{2(k+1)}}{2(x-4)^{2k}}\right|=|x-4|^2\lim_{k\to\infty}1=|x-4|^2<1[/tex]
[tex]\implies\boxed{3<x<5}[/tex]
b. ... [tex]f'(x)[/tex] converges for
[tex]\displaystyle\lim_{k\to\infty}\left|\frac{4(k+1)(x-4)^{2(k+1)-1}}{4k(x-4)^{2k-1}}\right|=|x-4|^2\lim_{k\to\infty}\frac{k+1}k=|x-4|^2<1[/tex]
[tex]\implies\boxed{3<x<5}[/tex]
c. ... [tex]f''(x)[/tex] converges for
[tex]\displaystyle\lim_{k\to\infty}\left|\frac{4(k+1)(2(k+1)-1)(x-4)^{2(k+1)-2}}{4k(2k-1)(x-4)^{2k-2}}\right|=|x-4|\lim_{k\to\infty}\frac{(k+1)(2k+1)}{k(2k-1)}=|x-4|^2<1[/tex]
[tex]\implies\boxed{3<x<5}[/tex]
d. ... [tex]\displaystyle\int f(x)\,\mathrm dx[/tex] converges to
[tex]\displaystyle\lim_{k\to\infty}\left|\frac{2\frac{(x-4)}^{2(k+1)+1}}{2(k+1)+1}}{2\frac{(x-4)}^{2k+1}}{2k+1}}\right|=|x-4|\lim_{k\to\infty}\frac{2k+3}{2k+1}=|x-4|<1[/tex]
[tex]\implies\boxed{3<x<5}[/tex]
