# How do you find a power series representation for #f(x)=ln(1+x)# and what is the radius of convergence?

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To find the power series representation for (f(x) = \ln(1+x)), we can use the known power series expansion for (\ln(1+x)):

[ \ln(1 + x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} ]

The radius of convergence for this power series can be found using the ratio test. The ratio test states that if ( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L ), then the radius of convergence is ( R = \frac{1}{L} ), where ( L ) is the limit.

Applying the ratio test to the power series for ( \ln(1 + x) ):

[ \lim_{n \to \infty} \left| \frac{(-1)^{n+2} \frac{(x^{n+1})}{n+1}}{(-1)^{n+1} \frac{x^n}{n}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{-x^{n+1} (n)}{(n+1)(x^n)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{-x}{n+1} \right| ]

[ = 0 ]

Thus, the limit of the absolute value of the ratio of consecutive terms is 0, indicating that the radius of convergence ( R ) is infinite. Therefore, the power series representation for ( f(x) = \ln(1+x) ) converges for all real numbers ( x ).

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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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