# 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)#?

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

# " " = 1 + x + 2/3x^2 + 3/8x^3 + 24/125x^4 + ... #

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

# " " = 1/2x^2+ 1/6x^3 +1/18x^4+ 3/360x^5+... #

The radii of convergence for the original and subsequent series is

# |x| < 1 #

We have:

Let us also define the following functions:

Then:

And:

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

Suppose that;

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

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

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To find the power series for ( f'(x) ), differentiate each term of the given series ( f(x) ) term by term and obtain the resulting series. To find the power series for ( \int f(t) , dt ) from ( 0 ) to ( x ), integrate each term of the series for ( f(x) ) term by term and add the constant of integration accordingly.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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