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
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)#