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

1 Answer
Nov 15, 2017

# f'(x) = sum_(n=1)^oo \ (n+1)x^(n-1) #

# int_0^x \ f(t) \ dt = sum_(n=1)^oo \ x^(n+1)/n #

Explanation:

We have:

# f(x) = sum_(1=0)^oo \ (n+1)/nx^n #

And so:

# f'(x) = d/dx sum_(n=1)^oo \ (n+1)/n \ x^n #

# " " = sum_(n=1)^oo \ d/dx \ (n+1)/n \ x^n #

# " " = sum_(n=1)^oo \ (n+1)/n \ nx^(n-1) #

# " " = sum_(n=1)^oo \ (n+1)x^(n-1) #

And:

# int_0^x \ f(t) \ dt = int_0^x \ sum_(n=1)^oo \ (n+1)/nt^n \ dt#
# " "= sum_(n=1)^oo \ (n+1)/n \ int_0^x \ t^n \ dt#

# " "= sum_(n=1)^oo \ (n+1)/n \ { \ [ t^(n+1)/(n+1)]_0^x } #

# " "= sum_(n=1)^oo \ (n+1)/n \ { \ x^(n+1)/(n+1)- 0 } #

# " "= sum_(n=1)^oo \ (n+1)/n \ x^(n+1)/(n+1) #

# " "= sum_(n=1)^oo \ x^(n+1)/n #