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)?

1 Answer
Andrea S.
Dec 21, 2017

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)

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