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)#?

Question

Luke Alvarez · Accepted Answer

# 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 # 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 #

Aurora Alvarado · Answer

undefined To find the power series for $ f'(x) $ and $ \int f(t) \, dt $ from $ [0,x] $ given the function $ f(x) = \sum_{n=1}^{\infty} \frac{n+1}{nx^n} $ from $ n=[1,\infty) $:

1. Differentiate the given function term by term to find $ f'(x) $.
2. Integrate the given function term by term to find $ \int f(t) \, dt $ from $ [0,x] $.