# How do you determine if the summation #n^n/(3^(1+2n))# from 1 to infinity is convergent or divergent?

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To determine if the series ( \sum_{n=1}^{\infty} \frac{n^n}{3^{1+2n}} ) is convergent or divergent, we can use the Ratio Test.

- Take the ratio of consecutive terms:

[ \frac{a_{n+1}}{a_n} = \frac{\left(\frac{(n+1)^{n+1}}{3^{1+2(n+1)}}\right)}{\left(\frac{n^n}{3^{1+2n}}\right)} ]

- Simplify the expression:

[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{3^{1+2(n+1)}} \cdot \frac{3^{1+2n}}{n^n} ]

[ = \frac{(n+1)^{n+1}}{3^{1+2n}} \cdot \frac{3^{1+2n}}{n^n} ]

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

- Simplify further using the properties of exponents:

[ = \left(\frac{n+1}{n}\right)^{n+1} ]

[ = \left(1 + \frac{1}{n}\right)^{n+1} ]

- Evaluate the limit as ( n ) approaches infinity:

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

- Use the fact that ( \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e ):

[ = e ]

- Analyze the result:

If the ratio ( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} ) 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.

Since the limit is ( e ), which is greater than 1, the series diverges by the Ratio Test. Therefore, the series ( \sum_{n=1}^{\infty} \frac{n^n}{3^{1+2n}} ) is divergent.

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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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