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
Steve M
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

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