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_{n}/a

_{n+1}|, I get for the Taylor series expansion of ln(x) around a=2 the answer of an infinite radius of convergence, which would mean that it would be valid everywhere, which would not make sense. What am I doing wrong?

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- Thread starter nomadreid
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In summary, the conversation discusses the calculation for the radius of convergence of a series and how it relates to the Taylor series expansion of ln(x) around a=2. The person is concerned about getting an infinite radius of convergence, which would mean the series is valid everywhere. However, they later realize their error and apologize for the misunderstanding.

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nomadreid said:What am I doing wrong?

We can't answer that unless you write your calculations.

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The radius of convergence for ln(x) around a=2 is 2. This means that the series will converge for all values of x within a distance of 2 from the center, which is a=2.

The radius of convergence for ln(x) is determined by first finding the interval of convergence, which is the distance from the center that the series will converge. This interval is then divided by 2 to get the radius of convergence.

No, the radius of convergence for ln(x) cannot be negative. The radius of convergence is always a positive value, as it represents a distance from the center.

When x is outside of the radius of convergence for ln(x), the series will not converge. This means that the series will either diverge or oscillate, depending on the specific values of x.

The value of a has a direct effect on the radius of convergence for ln(x). As a increases, the radius of convergence will decrease, and as a decreases, the radius of convergence will increase.

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