# How do you find the interval of convergence #Sigma (x-5)^n/(n!)# from #n=[0,oo)#?

The series:

is convergent for any

Apply the ratio test:

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The interval of convergence for the series ( \sum_{n=0}^{\infty} \frac{(x-5)^n}{n!} ) can be found using the Ratio Test.

By applying the Ratio Test, you take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term. If this limit is less than 1, the series converges. If it's greater than 1, the series diverges. If it's equal to 1, the test is inconclusive.

Applying the Ratio Test to ( \frac{(x-5)^{n+1}}{(n+1)!} \div \frac{(x-5)^n}{n!} ), simplifying, and taking the limit yields:

[ \lim_{n \to \infty} \frac{|x-5|}{n+1} ]

For convergence, this limit must be less than 1. Thus, we have ( |x-5| < \infty ), which implies ( -\infty < x - 5 < \infty ). Simplifying, we get ( -\infty < x < \infty ).

Therefore, the interval of convergence for the series is ( (-\infty, \infty) ).

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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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