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!)/n^nx^n# from #n=[1,oo)#?

The series for #f'(x)# is :

#f'(x)=sum_1^(oo) d/dx ((n!)/n^nx^n)#

#=sum_1^(oo) (n!)/n^(n-1)x^(n-1)#

From the FTC we know that:

#d/dx ( int_0^x f(t)dt) = f(x)#

It follows that the series for # int_0^x f(t)dt# is:

# int_0^x f(t)dt=sum_1^(oo)( int (n!)/n^nx^n dx)#

#=sum_1^(oo) (n!)/((n+1)n^n) x^(n+1) +C#

Because # int_0^0 f(t)dt = 0#, we conclude that #C = 0#

We have:

# f(x) = sum_(n=1)^oo \ (n!)/n^n \ x^n #
# " " = x + 1/2x^2 + 2/9x^3 + 3/32x^4 + 24/625x^5 + ... #

Let us also define the following functions:

# D(x) = f'(x) # and #I(x) = int_0^x \ f(t) \ dt #

Then:

# D(x) = d/dx sum_(n=1)^oo \ (n!)/n^n \ x^n #
# " " = sum_(n=1)^oo (n!)/n^n \ d/dx \ x^n #
# " " = sum_(n=1)^oo (n!)/n^n \ nx^(n-1) #
# " " = sum_(n=1)^oo (n!)/n^(n-1) \ x^(n-1) #
# " " = 1 + x + 2/3x^2 + 3/8x^3 + 24/125x^4 + ... #

And:

# I(x) = int_0^x \ sum_(n=1)^oo \ (n!)/n^n \ t^n \ dt #
# " " = sum_(n=1)^oo \ (n!)/n^n \ int_0^x \ \ t^n \ dt #
# " " = sum_(n=1)^oo \ (n!)/n^n \ [t^(n+1)/(n+1)]_0^x #

# " " = sum_(n=1)^oo \ (n!)/n^n \ x^(n+1)/(n+1) #
# " " = 1/2x^2+ 1/6x^3 +1/18x^4+ 3/360x^5+... #

For the radii of convergence, we can apply the d'Alembert's ratio test:

Suppose that;

# S=sum_(r=1)^oo a_n \ \ #, and #\ \ L=lim_(n rarr oo) |a_(n+1)/a_n| #

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

For the original series, the test limit is

# L = lim_(n rarr oo) | { ((n+1)!)/(n+1)^(n+1) \ x^(n+1) } / {(n!)/n^n \ x^n } | #
# \ \ = lim_(n rarr oo) | ((n+1)!)/(n+1)^(n+1) * n^n/(n!) * x^(n+1) / x^n | #
# \ \ = lim_(n rarr oo) | ((n+1))/(n+1)^(n+1) * n^n * x | #
# \ \ = lim_(n rarr oo) | 1/(n+1)^n * n^n * x | #
# \ \ = lim_(n rarr oo) | (n/(n+1))^n * x | #
# \ \ = lim_(n rarr oo) | (n/(n+1) * (1/n)/(1/n))^n * x | #
# \ \ = lim_(n rarr oo) | (1/(1+1/n))^n * x | #
# \ \ = | x | #

And so the series for #f(x)# is convergent if # |x| < 1 #

So for #f'(x)#, the test limit is:

# L = lim_(n rarr oo) | { ((n+1)!)/(n+1)^n \ x^n } / { (n!)/n^(n-1) \ x^(n-1) } | #
# \ \ = lim_(n rarr oo) | ((n+1)!)/(n+1)^n * n^(n-1)/(n!) * x^n/x^(n-1) | #
# \ \ = lim_(n rarr oo) | (n)/(n+1)^n * n^(n-1) * x | #
# \ \ = lim_(n rarr oo) | (n^n)/(n+1)^n * x | #
# \ \ = lim_(n rarr oo) | (n/(n+1))^n * x | #
# \ \ = lim_(n rarr oo) | (n/(n+1) * (1/n)/(1/n))^n * x | #
# \ \ = lim_(n rarr oo) | (1/(1+1/n))^n * x | #
# \ \ = | x | #

And so the series for #f'(x)# is convergent if # |x| < 1 #

And finally, for #I(x)#, the test limit is:

# L = lim_(n rarr oo) | { ((n+1)!)/ (n+1)^ (n+1) \ x^(n+2)/(n+2)} / {(n!)/n^n x^(n+1)/(n+1)} | #
# \ \ = lim_(n rarr oo) | ((n+1)!)/ (n+1)^ (n+1) * n^n/(n!) * x^(n+2)/(n+2) *(n+1)/x^(n+1)#
# \ \ = lim_(n rarr oo) | ((n+1))/ (n+1)^ (n+1) * n^n * x/(n+2) *(n+1)#
# \ \ = lim_(n rarr oo) | (1)/ (n+1)^n * n^n * x/(n+2) *(n+1) |#
# \ \ = lim_(n rarr oo) | (n/(n+1))^n * (n+1)/(n+2) * x | #
# \ \ = lim_(n rarr oo) | (n/(n+1) * (1/n)/(1/n))^n * (n+1)/(n+2) * (1/n)/(1/n)* x | #
# \ \ = lim_(n rarr oo) | (1/(1+1/n))^n * (1+1/n)/(1+2/n) * x | #
# \ \ = \ x | #

And so the series for #I(x)# is convergent if # |x| < 1 #

This should not come as any surprise as the radii of convergence is unaffected by differentiation or integration.