Question

How do you find the power series for `f'(x)` and `int f(t)dt` from [0,x] given the function `f(x)=Sigma sqrtnsqrt(n+1)x^n` from `n=[1,oo)`?

Answer 1

`f'(x) = sum_(n=1)^oo nsqrtnsqrt(n+1)x^(n-1)`

`int_0^xf(t)dt = sum_(n=1)^oo sqrtn/sqrt(n+1)x^(n+1)`

for `x in (-1,1)`

Explanation:

Consider for `x in RR` the series:

`sum_(n=1)^oo sqrtnsqrt(n+1)x^n`

using the ratio test we have:

`abs(a_(n+1)/a_n) = abs((sqrt(n+1)sqrt(n+2)x^(n+1))/(sqrtnsqrt(n+1)x^n)) = (sqrt(n+1)sqrt(n+2))/(sqrtnsqrt(n+1))absx = absx sqrt(1+2/n)`

so:

`lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) absx sqrt(1+2/n) = absx`

and we can conclude that the series is convergent for `x in (-1,1)`.
Inside the interval of convergence we can integrate and differentiate the series term by term, so if:

`f(x) = sum_(n=1)^oo sqrtnsqrt(n+1)x^n`

then:

`f'(x) = sum_(n=1)^oo nsqrtnsqrt(n+1)x^(n-1)`

`int_0^xf(t)dt = sum_(n=1)^oo sqrtn/sqrt(n+1)x^(n+1)`