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

Question

Christopher Agresta · Accepted Answer

The terms of this series are all positive, so the series is regular, that means that is divergent or convergent.
Since #n^n# is an infinite of higher order of #3^(1+2n)#, #lim_(nrarr+oo)n^n/3^(1+2n)=+oo#, the necessary condition (it has to be #0#) failed, so it is divergent.

Aurora Alvarado · Answer

undefined 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.

1. Take the ratio of consecutive terms:

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

2. Simplify the expression:

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

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

$$ = \frac{(n+1)^{n+1}}{n^n} $$

3. Simplify further using the properties of exponents:

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

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

4. Evaluate the limit as $ n $ approaches infinity:

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

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

$$ = e $$

6. 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.