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

The Taylor series for

Doing this gives:

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To find a power series representation for (\ln(1-x)), we can start with the known Taylor series expansion for (\ln(1+x)):

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

Substitute (-x) for (x) in the above series:

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

Simplify the series:

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

This is the power series representation for (\ln(1-x)).

The radius of convergence for this series can be found using the Ratio Test:

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

where (a_n) is the coefficient of (x^n) term in the series. In this case, (a_n = \frac{1}{n}).

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

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

[R = 1]

So, the radius of convergence for the power series representation of (\ln(1-x)) is (1).

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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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