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How do you apply the ratio test to determine if #sum_(n=1)^oo 3^n# is convergent to divergent?

Answer 1

Isabella Ackerman

The series diverges.

The the #n#th term in the series is given by #a_n=3^n.#

The ratio test states that the series should be convergent if:

#lim_(n->oo)a^(n+1)/a^n<1#

So, in our case we have:

#lim_(n->oo)3^(n+1)/3^n=lim_(n->oo)3^(n+1-n)=#

#=lim_(n->oo)3=3#

#3gt1# so the ratio test tells us this series diverges.

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Answer 2

Ezra Adams

To apply the ratio test to determine if the series $\sum_{n=1}^\infty 3^n$ is convergent or divergent:

Compute the ratio of consecutive terms: $\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{3^n} = 3$ .
Take the limit as $n$ approaches infinity: $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 3$ .
Analyze the limit:
- If the limit is less than 1, the series converges.
- If the limit is greater than 1 or does not exist, the series diverges.

Since the limit of $\frac{a_{n+1}}{a_n}$ is 3, which is greater than 1, the series $\sum_{n=1}^\infty 3^n$ diverges.

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Answer from HIX Tutor

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.