# How do you apply the ratio test to determine if #Sigma 2^n/(n!)# from #n=[1,oo)# is convergent to divergent?

The series is convergent and:

The ratio test states that a sufficient condition for a series:

to converge absolutely is that:

We then evaluate the ratio for the series:

so that:

proving that the series is convergent.

We should note by the way that the sum is:

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To apply the ratio test to determine the convergence or divergence of the series (\sum_{n=1}^{\infty} \frac{2^n}{n!}), we evaluate the limit of the absolute value of the ratio of successive terms as (n) approaches infinity:

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

where (a_n = \frac{2^n}{n!}).

First, we find (a_{n+1}) and (a_n):

[ a_{n+1} = \frac{2^{n+1}}{(n+1)!} = \frac{2 \cdot 2^n}{(n+1)n!} = \frac{2}{n+1} \cdot \frac{2^n}{n!} ]

[ a_n = \frac{2^n}{n!} ]

Next, we take the ratio of these terms:

[ \frac{a_{n+1}}{a_n} = \frac{\frac{2}{n+1} \cdot \frac{2^n}{n!}}{\frac{2^n}{n!}} = \frac{2}{n+1} ]

Finally, we find the limit of this ratio as (n) approaches infinity:

[ L = \lim_{n \to \infty} \frac{2}{n+1} = 0 ]

Since the limit of the ratio is less than 1, by the ratio test, the series (\sum_{n=1}^{\infty} \frac{2^n}{n!}) converges.

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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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