# How do you find the power series for #f'(x)# and #int f(t)dt# from [0,x] given the function #f(x)=Sigma x^n/lnn# from #n=[2,oo)#?

See below

For:

We differentiate wrt x:

And:

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To find the power series for ( f'(x) ) and ( \int f(t) , dt ) from ( [0,x] ) given the function ( f(x) = \sum_{n=2}^{\infty} \frac{x^n}{\ln n} ), we differentiate and integrate the given function term by term.

- To find ( f'(x) ): Differentiate each term of the series ( f(x) ) term by term to get ( f'(x) ).

[ f'(x) = \sum_{n=2}^{\infty} nx^{n-1} ]

- To find ( \int f(t) , dt ): Integrate each term of the series ( f(x) ) term by term with respect to ( t ) from ( 0 ) to ( x ) to get ( \int f(t) , dt ).

[ \int f(t) , dt = \sum_{n=2}^{\infty} \frac{x^{n+1}}{(n+1) \ln n} ]

So, the power series for ( f'(x) ) is ( \sum_{n=2}^{\infty} nx^{n-1} ), and the power series for ( \int f(t) , dt ) is ( \sum_{n=2}^{\infty} \frac{x^{n+1}}{(n+1) \ln n} ).

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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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