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

For

is convergent if the sequence:

is also convergent, and in such case:

Given the sequence:

Using the properties of logarithms:

that we can also write as:

Now use the limit:

which implies that:

and we have that:

If this last limit exists and is finite, i.e. if:

Then, as the exponential and the logarithm are continuous functions:

and:

is convergent if the sequence:

is also convergent, and in such case:

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See below.

Assuming that the formulation is

so

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To determine if the sequence ( a_n(1 + \frac{1}{\sqrt{n}})^n ) converges and find the limits when they exist, you can use the following steps:

- First, let's rewrite the sequence as ( a_n \left(1 + \frac{1}{\sqrt{n}}\right)^n ).
- Then, consider taking the limit as ( n ) approaches infinity: ( \lim_{n \to \infty} a_n \left(1 + \frac{1}{\sqrt{n}}\right)^n ).
- Apply the limit rules to simplify the expression and evaluate the limit.

If the limit exists and is finite, then the sequence converges. Otherwise, if the limit does not exist or is infinite, the sequence diverges.

Note: The specific value of ( a_n ) is not provided in the question, so the convergence or divergence of the sequence depends on the behavior of ( a_n ) and the term ( \left(1 + \frac{1}{\sqrt{n}}\right)^n ) as ( n ) approaches 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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