Question

Find the intervals of convergence of f(x), f '(x), f ''(x), and f(x) dx. (Enter your answer using interval notation. Be sure to check for convergence at the endpoints of the interval.) f(x) = ∞ (−1)n + 1(x − 4)x = 1 (a) f(x)

Answer 1

I'm guessing you're given

`f(x)=\displaystyle\sum_(n=1)^\infty((-1)^n+1)(x-4)^n`

Note that for odd
`n`, the corresponding term in the series is 0, so only the even terms matter. Let
`n=2k`, for which
`(-1)^(2k)+1=1+1=2`. Then

`f(x)=\displaystyle2\sum_(k=1)^\infty(x-4)^(2k)`

Differentiating/integrating the power series gives

`f'(x)=\displaystyle4\sum_(k=1)^\infty k(x-4)^(2k-1)`

`f''(x)=\displaystyle4\sum_(k=1)^\infty k(2k-1)(x-4)^(2k-2)`

`\displaystyle\int f(x)\,\mathrm dx=2\sum_(k=1)^\infty\frac{(x-4)}^(2k+1)}{2k+1}+C`

By the ratio test, ...

a. ...
`f(x)` converges for

`\displaystyle\lim_(k\to\infty)\left|(2(x-4)^(2(k+1)))/(2(x-4)^(2k))\right|=|x-4|^2\lim_(k\to\infty)1=|x-4|^2<1`

`\implies\boxed{3<x<5}`

b. ...
`f'(x)` converges for

`\displaystyle\lim_(k\to\infty)\left|(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`

`\implies\boxed{3<x<5}`

c. ...
`f''(x)` converges for

`\displaystyle\lim_(k\to\infty)\left|(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)((k+1)(2k+1))/(k(2k-1))=|x-4|^2<1`

`\implies\boxed{3<x<5}`

d. ...
`\displaystyle\int f(x)\,\mathrm dx` converges to

`\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)(2k+3)/(2k+1)=|x-4|<1`

`\implies\boxed{3<x<5}`