# How do you determine if #a_n=(1+n)^(1/n)# converge and find the limits when they exist?

The series converge to

Taking logarithms

Therefore,

So,

Apply the limit chain rule

Therefore,

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To determine if the sequence (a_n = (1+n)^{\frac{1}{n}}) converges, we can use the limit comparison test with the sequence (b_n = \frac{n}{n}).

Let's calculate the limit of the ratio (\frac{a_n}{b_n}):

[ \lim_{{n \to \infty}} \frac{(1+n)^{\frac{1}{n}}}{n^{\frac{1}{n}}} ]

Applying L'Hôpital's rule:

[ \lim_{{n \to \infty}} \frac{\frac{d}{dn}((1+n)^{\frac{1}{n}})}{\frac{d}{dn}(n^{\frac{1}{n}})} ]

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

[ = \lim_{{n \to \infty}} \frac{(1+n)^{\frac{1}{n}}}{n^{\frac{1}{n}}} ]

This limit is equal to 1. Therefore, (a_n) and (b_n) have the same behavior as (n \rightarrow \infty). Since (b_n) diverges (it's just (1)), (a_n) also diverges.

So, (a_n) diverges as (n) goes to infinity.

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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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